Published for SISSA by Springer2016)165.pdf · JHEP03(2016)165 Contents 1 Introduction1 2...

JHEP03(2016)165

Published for SISSA by Springer

Received December 15 2015

Accepted March 11 2016

Published March 23 2016

Interactions as intertwiners in 4D QFT

Robert de Mello Kocha and Sanjaye Ramgoolamb

aNational Institute for Theoretical Physics

School of Physics and Mandelstam Institute for Theoretical Physics University of Witwatersrand

Wits 2050 South AfricabCentre for Research in String Theory School of Physics and Astronomy

Queen Mary University of London

Mile End Road London E1 4NS UK

E-mail robertneophyswitsacza sramgoolamqmulacuk

Abstract In a recent paper we showed that the correlators of free scalar field theory in

four dimensions can be constructed from a two dimensional topological field theory based

on so(4 2) equivariant maps (intertwiners) The free field result along with recent results of

Frenkel and Libine on equivariance properties of Feynman integrals are developed further

in this paper We show that the coefficient of the log term in the 1-loop 4-point conformal

integral is a projector in the tensor product of so(4 2) representations We also show

that the 1-loop 4-point integral can be written as a sum of four terms each associated

with the quantum equation of motion for one of the four external legs The quantum

equation of motion is shown to be related to equivariant maps involving indecomposable

representations of so(4 2) a phenomenon which illuminates multiplet recombination The

harmonic expansion method for Feynman integrals is a powerful tool for arriving at these

results The generalization to other interactions and higher loops is discussed

Keywords AdS-CFT Correspondence Conformal and W Symmetry Duality in Gauge

Field Theories Gauge-gravity correspondence

ArXiv ePrint 151200652

Open Access ccopy The Authors

Article funded by SCOAP3doi101007JHEP03(2016)165

JHEP03(2016)165

Contents

1 Introduction 1

2 Background and motivations 5

21 CFT4TFT2 suggests equivariant interpretation of perturbative Feynman

integrals 5

22 Conformal integral exact answer and an expansion 6

23 Indecomposable representations and multiplet recombination 8

3 Harmonic expansion method and the logarithmic term 9

4 Coefficient of the log term and the projector 13

41 Analytic consequences 17

5 Quantum equation of motion indecomposable representations and equiv-

ariant maps 17

51 Quantum equations of motion 18

52 QEOM equivariant maps and their lifts 19

53 QEOM and twisted equivariant map 21

6 Conclusions and future directions 22

61 Towards higher loops 23

62 TFT2 and renormalization 24

63 Conformal blocks and CFT4TFT2 24

64 HPEM and the interactionintertwiner connection 25

A Basic formulae for HPEM (harmonic polynomial expansion method) 26

B Expansion of projector using the exact answer 28

B1 A summation formula for products of su(2) Clebschs from Feynman integrals 32

C Equivariant maps related to quantum equations of motion 33

C1 Quantum equation of motion for x1 33


1 Introduction

Many aspects of the combinatorics of N = 4 super-Yang-Mills theories have been shown

to be captured by two-dimensional topological field theories (TFT2s) based on permu-

tation groups [1ndash4 6ndash10] Specifically these topological field theories were found in the

ndash 1 ndash

JHEP03(2016)165

computation of correlators in the free limit of gauge theories the enumeration of states for

open strings connecting branes and the construction of their wavefunctions diagonalizing

the 1-loop dilatation operator [11 12] the enumeration of Feynman diagrams and tensor

model observables In the context of N = 4 SYM correlators this leads naturally to the

question of whether the space-time dependences of correlators (as well as the combina-

toric dependences on the operator insertions) can be captured by an appropriate TFT2

As a simple test case to explore this question we showed that free scalar field correla-

tors in four dimensions can be reproduced by a TFT2 with so(4 2) invariance [13] We

used Atiyahrsquos axiomatic framework for TFT2s where tensor products of a state space are

associated with disjoint unions of circles and linear homomorphisms are associated with

interpolating surfaces (cobordisms) [14] The properties of cobordisms in two dimensions

are reflected in the algebraic structure of a Frobenius algebra an associative algebra with

a non-degenerate pairing The notion of a TFT2 with global G symmetry was given in [15]

the state spaces are representations of the group G and the linear maps are equivariant

with respect to the G-action

In the construction of [13] the basic two-point function in scalar field theory is related

to the invariant map (V otimes V )rarr C where V is a direct sum of two irreducible representa-

tions of so(4 2) The irreducible representation (irrep) V+ contains a lowest weight state

corresponding to the basic scalar field via the operator-state correspondence

v+ = Lim|x|rarr0φ(x)|0 gt (11)

Translation operators ( Pmicro ) act on the lowest weight state to generate a tower of states

They can be viewed as raising operators since

[DPmicro] = Pmicro (12)

The state PmicroPmicrov+ is set to zero to correspond to the equation of motion of the scalar The

general state in this representation is

Y lm(P )v+ (13)

where Y lm(P ) is a symmetric traceless tensor of so(4) contracted with a product of P rsquos

The integer l gives the degree of the polynomial in P and m labels a state in the symmetric

traceless tensor representation We will refer to V+ as a positive energy representation

a terminology inspired by AdSCFT where the scaling dimension in CFT is energy for

global time translations in AdS [16ndash18] The irrep Vminus is dual to V+ It contains a dual

state vminus of dimension minus1 and other states are generated by acting with symmetric traceless

combinations of Kmicro

Y lm(K)vminus (14)

There is a non-degenerate invariant pairing η V+otimesVminus rarr C We refer to Vminus as a negative

energy representation since it contains states with negative dimension

ndash 2 ndash

JHEP03(2016)165

The foundation of the TFT2 approach to free CFT4 correlators is to consider the local

quantum field at x as a state in V+ oplus Vminus

Φ(x) =1radic2

(eminusiP middotxv+ + (xprime)2eiKmiddotx

primevminus)equiv Φ+ + Φminus (15)

Here xprimemicro = xmicro

x2 It is found that

η(Φ(x1)Φ(x2)) =1

(x1 minus x2)2(16)

We can think of eminusiP middotx and eiKmiddotxprime

as four dimensional analogs of the two dimensional vertex

operators familiar from string theory and 2D conformal field theory In 2D CFT physical

states of the string are constructed from exponentials of the coordinate quantum fields

Xmicro which have an expansion in oscillators coming from quantizing string motions In the

case of CFT4TFT2 at hand the exponential is in the momentum operators (and the

special conformal translations which are related to the momenta by inversion) which are

among the generators of so(4 2) Other developments in CFT4 inspired by vertex operators

include [19ndash22] It is intriguing that the 2D CFT vertex operators have the coordinate as

an operator in the exponential whereas here we are using the momenta as operators in

the exponential Conceivably there is some form of x minus p duality between these different

types of vertex operators Clarifying this could be useful in understanding the role of Born

reciprocity (a theme revived recently in [23 24]) in strings and QFT

The realization of CFT4 correlators in terms of TFT2 means that we are writing

quantum field theory correlators in terms of standard representation theory constructions

CFT4TFT2 builds on but goes beyond the standard use of representation theory as a

tool to calculate quantities defined by a path integral Rather it is a reformulation of

correlators of a quantum field theory in terms of standard constructions of representation

theory notably linear representations and equivariant maps between them The appearance

of both positive and negative energy representations in (15) is an important part of this

reformulation For example while the free field OPE

φ2 times φ2 rarr φ4 (17)

could be understood by using an expression for φ in terms of strictly positive energy

representations this is not the case for


The latter involves the invariant map η contracting a positive and a negative energy repre-

sentation This linearizes the CFT4 by relating correlators to linear equivariant maps The

construction achieves this by passing from the space of operators built on the primary at

x = 0 to the ldquodoubled spacerdquo of operators built on the primary at x = 0 (positive energy)

and x =infin (negative energy)

This paper addresses the natural question of whether the free field construction of [13]

is relevant to perturbative quantum field theory We explain this question in more tech-

nical terms in section 2 and show how it leads to the expectation that conformal integrals

ndash 3 ndash

JHEP03(2016)165

should be related to intertwiners involving representations of so(4 2) These integrals are

important building blocks in perturbation theory [26] and have been shown recently to have

remarkable properties called magic identities [27 28] Interestingly equivariance proper-

ties of the kind suggested by CFT4TFT2 have already been found in work of Frenkel and

Libine [29] who were approaching Feynman integrals from the point of view of quater-

nionic analysis Group-theoretic interpretations of relativistic holography have also been

suggested through the explicit construction of the boundary-to-bulk operators for arbitrary

integer spin as intertwining operators [30] The physics literature on higher dimensional

conformal blocks suggests equivariance properties of these integrals notably the works

of [31ndash33] which approach the conformal blocks in terms of Casimir differential equations

and subsequent reformulation in terms of the shadow formalism As indicated by the discus-

sion of OPEs above the QFT discussions of conformal blocks do not immediately imply an

interpretation in terms of linear representations and associated equivariant maps However

the use of Casimir differential equations is a powerful tool in arriving at the equivariant

map interpretation of QFT quantities The exponential vertex operators play an important

role in what follows because they allow us to map algebraic generators of the so(4 2) Lie

algebra to differential operators acting on function spaces In particular the Casimirs in

the (universal enveloping) so(4 2) algebra become Casimir differential operators

Section 2 reviews some aspects of the work of Frenkel-Libine which we will find useful

in developing the vertex operator approach to these equivariant maps We also review here

some basic facts about indecomposable representations which will be useful for section 5

In this paper our primary focus is on the conformal 4-point integral whose exact answer

is known [34 35] Our first main result is that the coefficient of the log-term in the 4-point

answer is given by the matrix elements of an equivariant map V+ otimes V+ otimes Vminus otimes Vminus rarr C

Section 3 reviews the harmonic expansion method which is used to arrive at this result

This method involves the expansion of the propagator in terms of so(4) harmonics For

a given order of the external points in the conformal integral (|x1| lt |x2| lt |x3| lt |x4|)we separate the integral into regions according to the range of integration of |x| One

region |x2| lt |x| lt |x3| leads to the logarithmic term The result that the coefficient of

the logarithmic term is an intertwiner is derived in section 4 This section contains our

first main result equation (415) Appendix B explains how the above result leads to an

identity for an infinite sum of products of su(2) Clebsch-Gordan coefficients

In section 5 we will consider the other regions of integration and show they can be

collected into four different terms associated with the quantum equation of motion for each

of the external variables xi On each of the terms the action of Laplacians gives so(4 2)

invariant equivariant maps involving a submodule of these indecomposable representations

For two of the four terms the equivariant maps employ the standard co-product and we

show how they can be lifted from the sub-module to the full indecomposable representation

The remaining two terms make use of a twisted co-product In these cases we believe the

lift to the full indecomposable representation is possible but there are technical subtleties

which remain to be clarified These results show that the full integral can be viewed as

an equivariant map obtained by lifting from the sub-module to the full indecomposable

representation Equation (523) is the second main result of this paper It links a beautiful

ndash 4 ndash

JHEP03(2016)165

structure in representation theory to quantum equations of motion arising from the collision

of interaction point with external points the source of many deep aspects of quantum field

theory The appearance of indecomposable representations is closely related to multiplet

recombination This phenomenon in connection with quantum equations of motion and

the Wilson-Fischer fixed point has also recently been discussed [36] Recombination of

superconformal multiplets has also been extensively discussed in the context of N = 4

and N = 2 theories (see for example [37ndash39] and refs therein) the breaking of higher spin

symmetry in AdSCFT being one of the motivations

In the final section we outline how our results extend to higher loops and describe other

future directions of research Notice that while we focus on the equivariance properties of

the integral to obtain a TFT we also require a complete characterization of the state space

a non-degenerate pairing and an associative product

2 Background and motivations


integrals

Once we have a formulation of all the correlators in free CFT4 in terms of TFT2 of equiv-

ariant maps the natural question is can we describe perturbation theory away from free

CFT4 in the language of the TFT2 Since perturbation theory involves the integration

of correlators in the free field theory weighted with appropriate powers of coupling con-

stants once we have a TFT2 description of all the free field correlators we are part of the

way there The important new ingredient is integration of the interaction vertices whose

consistency with equivariant maps remains to be established A natural place to start this

investigation is the case of conformal integrals [27 28] involving scalar fields It is known

that general perturbative integrals in four dimensions at one-loop can be reduced to a

basis of scalar integrals involving the box the triangle and bubble diagrams (see [26] and

refs therein) The momentum space box diagram becomes after Fourier transformation to

coordinate space a diagram related by graph duality to the original graph The integral

of interest in coordinate space is

I(x1 x2 x3 x4) =

intd4x

2π21

(x1 minus x)2(x2 minus x)2(x3 minus x)2(x4 minus x)2(21)

This integral (21) viewed as the kernel of an integral operator acting on appropriate test

functions has been shown to be related to equivariant maps in [29] There are two distinct

equivariant interpretations developed there one involves the Minkowski space integral and

the other involves integration over a U(2) in complexified space-time Subsequent higher

loop generalizations have been given [40 41]

Here we give a qualitative explanation of how the TFT2 way of thinking about per-

turbation theory suggests an equivariant interpretation for integrals Subsequently we will

investigate the expectations directly

We can choose all the external vertex operators to be

(xprime1)2eiKmiddotx

prime1vminus otimes (xprime2)

2eiKmiddotxprime2vminus otimes (xprime3)


2eiKmiddotxprime4vminus (22)

ndash 5 ndash

JHEP03(2016)165

Take a tensor product with

eminusiP middotxv+ otimes eminusiP middotxv+ otimes eminusiP middotxv+ otimes eminusiP middotxv+ (23)

Take a product of η pairings between the first factor in (22) with the first factor in (23)

the second with second etc This produces the product of propagators in (21) In another

way to set up the correlator use as external states

eminusiP middotx1v+ otimes eminusiP middotx2v+ otimes xprime23 eiKmiddotxprime3vminus otimes xprime24 eiKmiddotx

prime4vminus (24)

To this we tensor

(xprime)2eminusiKmiddotxprimevminus otimes (xprime)2eminusiKmiddotx

primevminus otimes eminusiP middotxv+ otimes eminusiP middotxv+ (25)

Again we pair the irsquoth factor in (24) with the corresponding factor in (25) All the

internal vertex operators have a common space-time position which is integrated over

The integrands can be reproduced by the TFT2 method

The different choices for external vertex operators should correspond to expansions in

positive powers of xi or of xprimei = xi|x2i |

A method of integration which connects with the above

vertex operator method of thinking about the integral is known as the Harmonic Polynomial

Expansion Method (HPEM) which give formulae that can be simplified using Gegenbauer

polynomials [42] We will choose an ordering of the external points |x1| lt |x2| lt |x3| lt |x4|and do the integral in Euclidean space separating it into five parts depending on the range

of |x| For each range we will apply the HPEM

The choice (24) corresponds to the region |x2| lt |x| lt |x3| where we will find a

logarithmic term There are no logs from any of the other regions This follows from basic

group theoretic properties of so(4) = su(2) times su(2) tensor products when these are used

in conjunction with the HPEM We will describe this in more detail in section 3 For now

we notice that the natural quantity to look at in search of an equivariant interpretation is

x23x24I(x1 x2 x3 x4) In section 4 we will establish that the coefficient of the log term in

x23x24I(x1 x2 x3 x4) can indeed be interpreted in terms of an equivariant map In arriving

at this we will make contact with the results of [29] in particular their discussion of a version

of the integral where the contour of integration is taken to be a copy of U(2) instead of

Minkowski space The relation between the coefficient of the log term and integration over

an alternative contour is reminiscent of discussions of generalized unitarity [43ndash45] Since

our discussion is in the framework of the HPEM in Euclidean space and involves a U(2)

integral it remains non-trivial to make direct contact with those discussions



of the external variables xi This separation will be used to give an interpretation in terms

of equivariant maps for the full integral

22 Conformal integral exact answer and an expansion

The integral (21) belongs to a class of conformal integrals which have been exactly solved

In momentum space the integral is a 1-loop box which can be computed in Minkwoski

ndash 6 ndash

JHEP03(2016)165

space using the usual iε prescription The result is [34 35]

I(x1 x2 x3 x4) =1

2x213x224

Φ(s t) (26)

where

Φ(s t) =1

λ

(2(Li2(minusρs) + Li2(minusρt)) + ln(ρs) ln(ρt) + ln

(t

s

)ln

(1 + ρt

1 + ρs

)+π2

3

)(27)

and

ρ =2

1minus sminus t+ λλ =

radic(1minus sminus t)2 minus 4st

s =x212x

234

x213x224

t =x214x

223

x213x224

(28)

We will need the expansion of Φ about s = 0 and t = 1 Towards this end we introduce

t = 1 + u and take the limit srarr 0 first and then urarr 0 In this limit

λ sim u+ s

(minus1minus 2

u

)+ s2

(minus 2

u3minus 2

u2

)sim u+ s

(minus1minus 2

u

)ρ sim s2

(2

u5+

1

u4

)+

s

u3+u2 minus usminus u2 + u+

1

uminus 1 sim u2 minus u

s(29)

In the limit we consider since ρrarrinfin we need to apply the identity

Li2(z) = minusLi2(1z)minus π2

6minus 1

2log2(minusz) (210)

to rewrite Li2(ρt) After this transformation

Φ(s t) =1

λ

2Li2(minusρs)minus 2Li2(minusρminus1tminus1)minus log

(st

)log

((1 + ρminus1tminus1)

(1 + ρs)

)(211)

Since we will discuss the coefficient of the log extensively in what follows we introduce the

notation

Φ(s u) = F0(s u) + log(s)F1(s u)

F0(s u) =1

λ

2Li2(minusρs)minus 2Li2(minusρminus1(1 + u)minus1) + log(1 + u) log


(1 + ρs)

)F1(s u) = minus 1

λlog


(1 + ρs)

)We are interested in the limit |x2| gt |x1| rarr 0 with |x4| gt |x3| rarr infin This means that

x1 x2 rarr 0 xprime3 xprime4 rarr 0 In this limit

s =(x1 minus x2)2(x3 minus x4)2

(x1 minus x3)2(x2 minus x4)2

=(x1 minus x2)2(xprime3 minus xprime4)2

f(x1 xprime3)f(x2 xprime4)

t =(x1 minus x4)2(x2 minus x3)2


=f(x1 x

prime4)f(x2 x

prime3)


ndash 7 ndash

JHEP03(2016)165

where

f(x yprime) = 1 + 2x middot yprime + x2yprime2 (212)

These equations show that u and s are real-analytic in the limit admitting expansions in

x1 x2 xprime3 xprime4 While λ ρ do not have an expansion in positive powers of s u as s u rarr

0 the quantity F1(s u) does have such an expansion This leads to an expansion of

F1(s(x1 x2 xprime3 xprime4) u(x1 x2 x

prime3 xprime4)) in powers of x1 x2 x

prime3 xprime4 will be related to a projector

in section 4

23 Indecomposable representations and multiplet recombination

We will review the notion of indecomposable representations and explain their relevance

to the recombination of multiplets when interactions are turned on

As a simple example consider the Lie algebra su(2) with generators J3 Jplusmn

[J3 Jplusmn] = plusmnJplusmn[J+ Jminus] = 2J3 (213)

With this normalization of the generators irreducible representations have J3 eigenvalues in

the range j jminus1 middot middot middot minusj for j isin 0 12 132 middot middot middot Consider a lowest weight representation

built by starting with a state | minus 12〉 satisfying

J3

∣∣∣minus 1

2

rang= minus1

2

∣∣∣minus 1

2

rangJminus

∣∣∣minus 1

2

rang= 0 (214)

Now consider the infinite dimensional representation spanned by Jn+| minus 12〉 for n isin

0 1 2 middot middot middot Denote this representation by Vminus 12 The state J2

+| minus 12〉 has the property

that it is annihilated by Jminus

JminusJ2+

∣∣∣minus1

2

rang= 0 (215)

This has the consequence that the vector subspace of Vminus 12

spanned by Jn+| minus 12 gt for

n ge 2 is an invariant subspace of Vminus 12 Denote this subspace as V

(2)

minus 12

The quotient

space Vminus 12V

(2)

minus 12

is the standard two-dimensional representation of su(2) We have an exact

sequence

0rarr V(2)

minus 12

rarr Vminus 12rarr Vminus 1

2V

(2)

minus 12

rarr 0 (216)

The quotient space admits a positive definite inner product If we choose an inner product

where | minus 12〉 has unit norm then J2

+| minus 12〉 has zero norm Setting this null state to zero

gives the quotient space which is a unitary representation of su(2)

In four dimensional free scalar quantum field theory we encounter the representation

V+ containing a lowest weight state v+ of dimension 1 There are additional states of

higher dimension of the form

Tmicro1micro2middotmiddotmiddotmicronI Pmicro1 middot middot middotPmicronv+ (217)

ndash 8 ndash

JHEP03(2016)165

where the TI are symmetric traceless tensors This is a unitary representation of so(4 2)

By direct analogy to the above discussion V+ is obtained as a quotient space of a larger

representation V+ spanned by

Smicro1middotmiddotmiddotmicronI Pmicro1 middot middot middotPmicronv+ (218)

where the SI are symmetric tensors (not necessarily traceless) To get to V+ we quotient

V+ by the subspace spanned by

Smicro1middotmiddotmiddotmicronI Pmicro1 middot middot middotPmicronPmicroPmicrov+ (219)

Denoting this subspace by V(p2)+ we have the exact sequence

0rarr V(p2)+ rarr V+ rarr V+ = V+V

(p2)+ rarr 0 (220)

The representation V+ is generated by acting with derivatives on the elementary scalar

field and using the operator-state correspondence The representation V(p2)+ is isomorphic

to the representation obtained by taking all derivatives of φ3 in free scalar field theory and

applying the operator-state correspondence When we perturb the free theory with a φ4

interaction we have the quantum equation of motion

partmicropartmicroφ = gφ3 (221)

This quantum equation of motion and its relation to the indecomposable representation

V+ is reflected in the properties of the integral (21) This will be the subject of section 5

Indecomposable representations have appeared in discussions of 2D CFT see for exam-

ple [46ndash48] Our observations draw some elements from this work eg in the use we make

of twisted co-products in connection with OPEs but they are not a direct translation of the

2D story which relies on the use of the complex coordinates (z z) and the corresponding

chiral-anti-chiral factorization

3 Harmonic expansion method and the logarithmic term

The harmonic expansion method expands the two point function in terms of products

of spherical harmonics In this way the action of so(4 2) on any of the four external

coordinates is manifest The form of the expansion is dictated by the relative sizes of the

integration variable and the external coordinates Consequently this expansion method

breaks the integration region down into a set of 5 regions The main result of this section

is an explicit answer for each of these regions This allows us to isolate the logarithmic

term to be discussed further in section 4 It also gives a neat separation of the integral

into terms which are homogeneous and inhomogeneous terms for each of the Laplacians i

which will be useful for the equivariant interpretation of the quantum equations of motion

in section 5

Let |x1|lt |x2|lt |x3|lt |x4| First consider the region where |x| is less than all the |xi|

I1 =sumllimi

prod4i=1 Y

limi(x

primei)

|x1|2|x2|2|x3|2|x4|2

int |x1|0

drr3rl1+l2+l3+l4intd3x

4prodi=1

Y mili

(x) (31)

ndash 9 ndash

JHEP03(2016)165

where d3x = 12π2dS

3 with dS3 the standard measure on the unit sphere The last factor

is a group theoretic factor which will appear in all of the five integration regions We can

write it as

Cm1m2m3m4

l1l2l3l4=

suml5m5m6

Cm1m2l5l1l2m5

1

(l5 + 1)Cm3m4l5l3l4m6

δ(m1 +m2m5)δ(m3 +m4m6)gm5m6

(32)

where

Y m1l1

Y m2l2

=suml5m5

Cm1m2l5l1l2m5

Y m5l5

(33)

This is the Clebsch-Gordan coefficient for multiplication of spherical harmonics on S3

Selection rules for C l1l2l5m1m2m5 imply that

max(l1 l2)minusmin(l1 l2) le l5 le l1 + l2 (34)

or equivalently

|l1 minus l2| le l5 le l1 + l2 (35)

If we multiply two symmetric traceless tensors T1 T2 of ranks l1 and l2 we can get some-

thing symmetric and traceless of rank l1 + l2 If we contract two indices one from each

we can reduce the rank by 2 Further such contractions reduce the rank by multiples of

2 And the maximum number of contractions is min(l1 l2) For the 4-point coupling of

spherical harmonics to be non zero we need

l4 isin l1 + l2 + l3 l1 + l2 + l3 minus 2 l1 + l2 + l3 minus 4 middot middot middot (36)

Alternatively a convenient way to parametrize the possibilities is given by

l1 + l2 minus 2k12 = l3 + l4 minus 2k34 (37)

where 0 le k12 le min(l1 l2) and 0 le k34 le min(l3 l4)

After doing the integral

I1 =sumlimi

prodi Y

limi(x

primei)

|x1|2|x2|2|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

|x1|4+l1+l2+l3+l44 + l1 + l2 + l3 + l4

(38)

We will write I1 = IS11 which indicates that the radial position of the interaction point

coincides with the radial position of |x| as we evaluate this integral The superscript

indicates that the answer is a power series in the xprimemicroi

ndash 10 ndash

JHEP03(2016)165

Next consider the region |x1| lt |x| lt |x2| The contribution to the integral from this

region is

I2 =sumlimi

Y l1m1

(x1)prodi=2 Y

limi(x

primei)

|x2|2|x3|2|x4|2

int |x2||x1|

dr r3 rminus2intd3xY m1

l1(xprime)Y m2

l2(x)Y m3

l3(x)Y m4

l4(x)

=sumlimi

Y l1m1

(x1)prodi=2 Y

limi(x

primei)

|x2|2|x3|2|x4|2

int |x2||x1|

dr r3 rminus2rminusl1+l2+l3+l4Cm1m2m3m4

l1l2l3l4

=sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

|x2|2Y l3m3

(xprime3)

|x3|2Y l4m4

(xprime4)

|x4|2Cm1m2m3m4

l1l2l3l4

(minusl1 + l2 + l3 + l4 + 2)

times(rminusl1+l2+l3+l4+22 minus rminusl1+l2+l3+l4+2

1

)(39)

We used x = xprime Y lm(xprime) = rminuslY l

m(x) and Y lm(x) = rlY l

m(x) Note that minusl1 + l2 + l3 + l4 ge 0

follows from the selection rules for su(2) tensor products Define IS21 IS22

IS21 =sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

|x2|2Y l3m3

(xprime3)

|x3|2Y l4m4

(xprime4)

|x4|2Cm1m2m3m4

l1l2l3l4

(minusl1 + l2 + l3 + l4 + 2)

(minusrminusl1+l2+l3+l4+2

1

)IS22 =

sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

|x2|2Y l3m3

(xprime3)

|x3|2Y l4m4

(xprime4)

|x4|2Cm1m2m3m4

l1l2l3l4

(minusl1 + l2 + l3 + l4 + 2)

(rminusl1+l2+l3+l4+22

)(310)

IS21 is obtained from the limit where the radial position of the integrated interaction point

coincides with the radial position of the external leg x1 ie where |x| = |x1| The super-

script indicates that this is a power series in the xmicro1 xprimemicro2 x

primemicro3 x

primemicro4 variables IS22 is analogously

defined in terms of |x| = |x2| We have

I2 = IS21 + IR21 (311)

Now consider the third region where |x| is in the middle

I3 =sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

int |x3||x2|

dr r3minus4+l3+l4minusl1minusl2

=sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4log

(|x3||x2|

)δ(l1 + l2 l3 + l4)

+sum

limil 6=0

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2

(|x3|l minus |x2|l

l

)δ(lminusl1 minus l2 + l3 + l4)

(312)

ndash 11 ndash

JHEP03(2016)165

It is convenient to define

IL32 =minussumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4log (|x2|) δ(l1 + l2 l3 + l4)

IS32 =sum

limil 6=0

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

(minus|x2|l

l


IL33 =sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4log (|x3|) δ(l1 + l2 l3 + l4)

IS33 =sum

limil 6=0

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

(|x3|l

l


(313)

IL32 is the logarithmic term coming from the |x| = |x2| end of the integral where the radial

position of the interaction point coincides with the radius of the external point x2 IS32 is

the series term from the same limit IL33 IS33 have been defined analogously Again

I3 = IL32 + IS32 + IL33 + IS33 (314)

The integer l appearing in (312) can be positive or negative If we assume x1 x2 are

small and x3 x4 large - we can specialize the known answers to the integral In that case

we know that

|x2| = max(r1 r2) =1

2(r1 + r2 + |r1 minus r2|)

|x3| = min(r3 r4) =1

2(r3 + r4 minus |r3 minus r4|) (315)

It is also useful to express the result in terms of rprime3 = 1r3 which gives

log (r2r3) = log((r1 + r2 + |r1 minus r2|)(rprime3 + rprime4 + |rprime3 minus rprime4|)

)(316)

Note that in the limit of r1 r2 sim εrarr 0 rprime3 rprime4 sim εrarr 0 this goes like log(ε2) just like log s

In section 4 we will give the precise relation between the coefficient of log s in the exact

answer (what we call F1(s t)) and the coefficient of log (r2r3) computed above

In the fourth region |x| is between |x3| and |x4|

I4 =sumlimi

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)

|x4|2C l1l2l3l4m1m2m3m4

int |x4||x3|

dr r3minus6rminusl1minusl2minusl3+l4

=sum

limille0

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)

|x4|2

(|x4|lminus2 minus |x3|lminus2

l minus 2

)δ(lminusl1 minus l2 minus l3 + l4) (317)

Note that there are no log terms here since l is never equal to 2 It is useful to define

I43 =sum

limille0

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)

|x4|2

(minus|x3|lminus2

l minus 2

)δ(lminusl1 minus l2 minus l3 + l4)

I44 =sum

limille0

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)

|x4|2

(|x4|lminus2

l minus 2

)(318)

ndash 12 ndash

JHEP03(2016)165

I43 is obtained from the lower limit where |x| = |x3| with the radial position of the

interaction point coinciding with the radial position of x3 I44 is obtained from the upper

limit |x| = |x4| and

I4 = I43 + I44 (319)

The fifth region is given by |x| gt |xi|

I5 =sumlimi

Cm1m2m3m4

l1l2l3l4

prodi

Y limi(xi)

int infin|x4|

drr3minus8rminusl1minusl2minusl3minusl4

=sumlimil

Cm1m2m3m4

l1l2l3l4

prodi

Y limi(xi)

(|x4|minus4minusl

4 + l

)δ(l l1 + l2 + l3 + l4) (320)

In this case there is no log term as the li are all integers greater than or equal to zero We

write I5 = IS54 to indicate that this is a power series expansion and arises from the integral

at the limit |x| = |x4|The integral I = I1 + I2 + I3 + I4 + I + 5 is a contribution to the four point function

of free scalar fields at points x1 x2 x3 x4 Each field has dimension ∆ = 1 and spin zero

Consequently acting with the quadratic Casimir on any field must give

C2 = ∆(∆minus 4) + l(l + 2) = minus3 (321)

In appendix A we explain how to translate C2 into a differential operator Using the

resulting differential operator (C2)i in any of the coordinates xi we verify that

(C2)iI(j) = minus3I(j) i j isin 1 2 3 4 (322)

4 Coefficient of the log term and the projector

We are computing I(x1 x2 x3 x4) with specified ordering |x1| lt |x2| lt |x3| lt |x4| Apply-

ing the HPEM there is a logarithmic term coming from the range |x2| le x le |x3| In this

section we want to argue that the coefficient of the logarithmic term has a representation

theory interpretation as an invariant map built from a projection operator P++++ that we

define below The projection operator P++++ featured prominently in the work of Frenkel

and Libine [29]

The logarithmic term coming from the HPEM was computed in the last section The

result is

log(r3r2

)sumlimi

(Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

)δ(l1 + l2 l3 + l4) (41)

The exact result for I was given in (26) in terms of

Φ(s t) = F0(s t) + log(s)F1(s t) (42)

Consider the Casimir

C2 = ηACηBD(L(1)AB + L(2)AB)(L(1)CD + L(2)CD) (43)

ndash 13 ndash

JHEP03(2016)165

of so(4 2) acting on the coordinates x1 x2 For any function H(s t) of the conformal cross

ratios the quadratic Casimir C2 of so(4 2) becomes the differential operator [49]

C2H = 2(1 + sminus t)st part2H

partsparttminus(1minus s+ t

)spart

parts

(spartH

parts

)minus(

(1minus t)2 minus s(1 + t)) partpartt

(tpartH

partt

)+ 4s

partH

parts(44)

Using the above differential operator we find

C2 sΦ = 2 sΦ C2 sF1 = 2 sF1 (45)

Thus the Casimir equation obeyed by the full integral is also obeyed by the coefficient of

the log term From (26) we see that the coefficient of log(s) in the known exact answer

for the integral is

1

2x213x224

F1(s t) (46)

The log(s) appearing in (26) is the only possible source of log r3r2

dependence which im-

plies that

2sumlimi

(Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

)δ(l1 + l2 l3 + l4) =

1

x213x224

F1(s t)

(47)

The representation V+ has lowest weight state of dimension 1 written as V+ In the

notation of Dolan [50] it is D[100] The tensor product V+ otimes V+ can be decomposed into

a direct sum of irreducible representations [50 51]

V+ otimes V+ = A[200] +

infinoplusk=1

D[k+2 k2 k2] (48)

Given such a decomposition of a tensor product into a direct sum there are projectors for

each of the terms These projectors commute with the so(4 2) actions and hence describe

equivariant maps The representation A[200] will henceforth be called V++ and corresponds

to the CFT primary operator φ2 and its descendants There are Clebsch-Gordan maps

M V+ otimes V+ rarr V++

Mdagger V++ rarr V+ otimes V+ (49)

which are equivariant maps between the tensor product and the irrep There is a projector

P++++ defined by

P++++ V+ otimes V+ rarr V+ otimes V+P++++ =MMdagger (410)

There is a closely related projector P++minusminus

P++minusminus V+ otimes V+ otimes Vminus otimes Vminus rarr C (411)

ndash 14 ndash

JHEP03(2016)165

Figure 1 The construction of P++minusminus follows by composing P and two invariant pairings (η)

This is obtained by tensoring both sides of (410) with Vminus otimes Vminus

P V+ otimes V+ otimes Vminus otimes Vminus rarr V+ otimes V+ otimes Vminus otimes VminusP = P 1VminusotimesVminus (412)

The rhs of the first line of (412) can be equivariantly mapped to C by using the invariant

pairing between the first V+ and first Vminus and the invariant pairing between second V+ and

second Vminus as illustrated in figure 1 Composing P with these invariant pairings gives

P++minusminus We can evaluate this projector on position eigenstates

P++minusminus(eminusiP middotx1v+ otimes eminusiP middotx2v+ otimes eiKmiddotxprime3vminus otimes eiKmiddotxprime4vminus) equiv P(x1 x2 xprime3 xprime4) (413)

We can also evaluate it on spherical harmonics

P++minusminus(Y l1m1

(P )v+ otimes Y l2m2


(K)vminus otimes Y l4m4

(K)vminus) equiv P l1l2l3l4m1m2m3m4(414)

Our claim is that the power series expansion of 12x213x

224F1 at small x1 x2 x

prime3 xprime4 coincides

with that of P(x1 x2 xprime3 xprime4)

1

2x213x224

F1(u(x1 x2 xprime3 xprime4) s(x1 x2 x

prime3 xprime4)) = xprime23 x

prime24 P(x1 x2 x

prime3 xprime4) (415)

This is the main result of this section This power series expansion can be conveniently

organised in terms of the coefficients P l1l2l3l4m1m2m3m4

Consider the coefficient of the product of harmonics in the log term (41) The spherical

harmonics Y lm(x) are a basis for harmonic functions regular at x = 0 and carry a repre-

sentation of so(4 2) Denote the corresponding function space H+ Similarly (xprime)2Y lm(xprime)

are a basis for harmonic functions regular at xprime = 0 ie x = infin They also carry a

representation of so(4 2) Denote the corresponding function space Hminus Picking up the

coefficient of the harmonics is mapping H+ otimesH+ otimesHminus otimesHminus rarr C This coefficient is just

ndash 15 ndash

JHEP03(2016)165

the tensor Cm1m2m3m4

l1l2l3l4defined in equations (32) and (33) in terms of the structure con-

stants for multiplication of spherical harmonics The 3-point structure constants involve

the integration intd3x Y l1

m1(x)Y l2

m2(x)Y l5

m5(x) (416)

Thus picking up the coefficient of Y l1m1

(x1)Yl2m2

(x2) involves mapping

Y l1m1

(x1)otimes Y l2m2

(x2) rarr rl1+l2Y l1m1

(x)Y l2m2

(x)

Y l3m3

(xprime3)

|x3|2otimesY l4m4

(xprime4)

|x4|2rarr

Y l3m3

(xprime)

|x|2otimesY l4m4

(xprime)

|x|2(417)

These are applications of the equivariant mapsM H+otimesH+ rarr H++ andM HminusotimesHminus rarrHminusminus as explained in section 52 of [29] From a physical perspective this corresponds to

the fact that the free scalar field φ(x) has modes transforming in H+ (and Hminus) while the

field φ2(x) has modes transforming in H++ ( and Hminusminus) After these maps are applied the

HPEM sets l1 + l2 = l + 3 + l4 when we do the radial integral and pick up the log term

There remains an integral over S3 which gives the factor (l5 + 1)minus1 This corresponds

in the discussion of [29] (proposition 84) (see also equation (12) of [40]) to an integral

over U(2) Thus we have a direct link between the integration over a modified contour in

complexified space-time (where we are integrating over U(2) instead of Minkowski space)

and the coefficient of the log term This is likely to be an example of a general story that

should hold for more general Feynman integrals

We can use the vertex operators of TFT2 to further clarify the discussion The natural

language for the above discussion is in terms of a map H+otimesH+otimesHminusotimesHminus rarr C Using the

vertex operators we will see that it is equally natural to employ a map V+otimesV+otimesVminusotimesVminus rarrC Start with (413) and expand the exponentials in Y l

m(P )Y ml (x) or Y l

m(K)Y ml (xprime)

sumlimi

(minusi)l12l1 l1

Y l1m1

(x1)(minusi)l22l2 l2

Y l2m2

(x2)(i)l3

2l3 l3Y l3m3

(xprime3)(i)l4

2l4 l4Y l4m4

(xprime4)

P++minusminus

(Y m1l1

(P )v+ otimes Y m2l2


(K)vminus otimes Y m4l4

(K)vminus)

=sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

P++minusminus

(Y m1l1

(P )

2l1 l1v+ otimes

Y m2l2

(P )

2l2 l2v+ otimes

Y m3l3

(K)

2l3 l3vminus otimes

Y m4l4

(K)

2l4 l4vminus)

(418)

The vertex operators of TFT2 provide equivariant maps between the algebraic state spaces

and the polynomial state spaces which makes it possible to express the projector in terms

of these state spaces Indeed the above argument makes it clear thatprodY limi(xi) multiplies

a projector acting on states in V+ otimes V+ otimes Vminus otimes Vminus This shows that the claim that the

series expansion multiplying the log is the evaluation of an so(4 2) invariant projection on

states created from the vplusmn by vertex operators is equivalent to saying that the coefficient

of the product of spherical harmonics has to be an equivariant map

ndash 16 ndash

JHEP03(2016)165

41 Analytic consequences

We have seen that the coefficient of the log term in the HPEM has an expansion in powers

of x1 x2 xprime3 xprime4 We will now see how the same expansion arises from the exact answer

We want to consider the limit s rarr 0 t rarr 1 for the conformal cross ratios It proves

to be useful to set t = 1 + u and then consider urarr 0 The coefficient of log s is

F1 =1

λlog

((1 + ρs)

(1 + ρminus1tminus1)

)(419)

Expanding using Mathematica we find

F1(s u) =infinsumk=0

sk

u2k+1[Qk(u) log(1 + u)]+ (420)

Here Qk(u) is a polynomial in u and [Qk(u) log(1 + u)]+ is the truncation of the power

series in u to powers un with n ge 2k + 1 Qk(u) is order k

Qk(u) =

ksuml=0

bl(k)ul (421)

where

bl(k) =

(k

k minus l

)(2k + l

k minus l

)(422)

Note that the existence of such an expansion is non-trivial The individual factors such

as ρ λ when expanded in positive powers of s contain at each order a finite number of

negative powers of u Nevertheless the combination of terms appearing in F1 is analytic

in u s at u s sim 0 Appendix B explains how we arrived at the above formula with the

help of Mathematica The appendix also explains how the discussion implies a summation

formula for products of su(2) Clebsch-Gordan coefficients in terms of F1

5 Quantum equation of motion indecomposable representations and

equivariant maps

Using the harmonic expansion method we have found

I = I1 + I2 + I3 + I4 + I5

I = (IS11 + IS21) + (IS22 + IS32 + IL32) + (IL33 + IS33 + IS43) + (IS45 + IS55) (51)

Rearrange these contributions by defining

I(1) = (IS11 + IS21)

I(2) = (IS22 + IS32 + IL32)

I(3) = (IL33 + IS33 + IS43)

I(4) = (IS45 + IS55) (52)

ndash 17 ndash

JHEP03(2016)165

This reorganization is such that I(i) arises from integration limits where the radial posi-

tion of the interaction point coincides with the irsquoth external coordinate In terms of the

quantities just introduced we have

I = I(1) + I(2) + I(3) + I(4) (53)

Due to radial ordering the order of the fields within the correlator swaps when moving

from one term to the next As a consequence of these discontinuities we expect that

jI(i) = minus2δij

prodk 6=j

1

(xk minus xi)2(54)

We will demonstrate using the explicit formulae from the HPEM that this is indeed the

case The I(i) are also used to develop equivariant map interpretations for the full integral

I Each I(i) is the starting point for one equivariant map interpretation We exhibit the

complete story for I(1) while the discussion for I(4) is related by inversion We outline the

story for I(2) (and by inversion for I(3)) It has an additional intricacy involving the use of

a twisted co-product This raises some technical problems which we leave for the future

51 Quantum equations of motion

Consider the term I(1) which is given by

I(1) =sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

r22

Y l3m3

(xprime3)

r23

Y l4m4

(xprime4)

r24Cm1m2m3m4l1l2l3l4

times (minus2)(l1 + 1)r2+l2+l3+l4minusl11

(l1 + l2 + l3 + l4 + 4)(minusl1 + l2 + l3 + l4 + 2)(55)

We have written the above formula in terms of a product of harmonic functions in

x1 xprime2 xprime3 xprime4 so that it has a smooth x1 rarr 0 limit as well as a smooth x2 x3 x4 rarr infin

limit To apply the Laplacian to the above result the formulas

part2

partxmicro1partxmicro1

(r21)A = 4A(A+ 1)(r21)Aminus1

part

partxmicro1(r21)A = 2A(r21)Aminus1xmicro1

part

partxmicro1Y l1m1

(x1)part

partxmicro1(r21)A = 2Al1(r

21)Aminus1Y l1

m1(x1) (56)

are useful It is now simple to obtain

part2


I(1) =sumlimi

minus2(l1 + 1)(r1)l2+l3+l4minusl1+2

r22r23r

24

Y l1m1

(x1)Yl2m2

(xprime2)Yl3m3

(xprime3)Yl4m4

(xprime4)Cm1m2m3m4l1l2l3l4

(57)

To recognize the right hand side note that

1

|x1minusx2|2|x1minusx3|2|x1minusx4|2=

1

r22r23r

24

sumlimi

Y l2m2

(x1)Yl3m3

(x1)Yl4m4

(x1)Yl2m2

(xprime2)Yl3m3

(xprime3)Yl4m4

(xprime4)

=(l1 + 1)rl2+l3+l41

r22r23r

24

sumlimi

Cm1m2m3m4l1l2l3l4

Y l1m1

(x)Y l2m2

(xprime2)Yl3m3

(xprime3)Yl4m4

(xprime4) (58)

ndash 18 ndash

JHEP03(2016)165

The (l1+1) in the numerator arises because of the normalization of the spherical harmonics

Clearly then we have demonstrated (54)

The harmonic expansion method expands the propagators in spherical harmonics which

solve Laplacersquos equation How then did we get a non-zero answer The point is that when

|x| lt |x1| we are expanding in positive powers of xprime1 and when |x| gt |x1| we are expanding

in positive powers of x1 In each case although the x1 dependent functions are harmonics

the integration produces an additional dependence on x1 from the integration limits In

the operator formalism where we compute a radially ordered correlator the ordering of the

interaction vertex changes relative to the external point x1 when we move from the region

|x| lt |x1| to the region |x| gt |x1| So as expected the violation of the free equation has to

do with collision of the integration point with an external coordinate

The contribution I(1) did not include a log dependence We will consider one more

example I(2) chosen because this term does include a log dependence

I(2) =sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


1

l2+l3+l4minusl1+2

δ(l1 + l2 l3 + l4)

r23r24

+sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


1

r23r24

times

[minus rminusl1minusl2+l3+l42 2(l2 + 1)

(l2+l3+l4minusl1+2)(l3+l4minusl1minusl2)(1minusδ(l1+l2 l3+l4))minus δ(l1+l2 l3+l4)log(r2)

]

We will again make use of the formulas above in (56) as well as

part

partxmicro2Y l2m2

(x2)part


22)Aminus1Y l2

m2(x2)

part

partxmicro2logr2 =

xmicro2r22

part

partxmicro2

part

partxmicro2logr2 =

2

r22(59)

We find

part2


I(2) = minus2sumlimi

(l2+1)rminusl1minusl2+l3+l4minus22

r23r24

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


(510)

The right hand side can again be identified with

minus 2

|x1 minus x2|2|x2 minus x3|2|x2 minus x4|2(511)

The log contributes the term with l1 + l2 = l3 + l4 in (510)

The discussion for the terms I(3) and I(4) is now straight forward

52 QEOM equivariant maps and their lifts

As we discussed there is an equivariant map between V(p2)+ and the irrep generated by the

field φ3 ie the irrep V+++ obtained by acting with (P middot middot middotP ) on v+ otimes v+ otimes v+ Given

ndash 19 ndash

JHEP03(2016)165

the TFT2 construction of free field correlators [13] we know that there is an so(4 2)

equivariant map

F1 V+++ otimes Vminus otimes Vminus otimes Vminus rarr C (512)

such that

F1(eminusiP middotx1v+++ otimes eiKmiddotx

prime2vminus otimes eiKmiddotxprime3vminus otimes eiKmiddotxprime4vminus) =

x22x23x

24

(x1minusx2)2(x1minusx3)2(x1minusx4)2(513)

For completenesswe give a derivation in appendix C Given the isomorphism between V(p2)+

and V+++ we have a map

F1 V(p2)+ otimes Vminus otimes Vminus otimes Vminus rarr C (514)

It is given similarly by

F1(eminusiP middotx1PmicroPmicrov

+ otimes eiKmiddotxprime2vminus otimes eiKmiddotxprime3vminus otimes eiKmiddotxprime4vminus) =x22x

23x

24

(x1 minus x2)2(x1 minus x3)2(x1 minus x4)2= f(x1 x

prime2)f(x1 x

prime3)f(x1 x

prime4)

(515)

The function f(x1 xprime2) is the series in positive powers of x1 x

prime2 which sums to

1

(1minus 2x1 middot xprime2 + x21xprime22 )

(516)

A consistency check of this interpretation is that the Casimirs for each of the four

so(4 2)rsquos one for each coordinate xi gives the value (minus3) appropriate for Vplusmn Vplusmn (322)

This map F1 can be lifted from the subspace V(p2)+ to the larger space V+

F1 V+ otimes Vminus otimes Vminus otimes Vminus rarr C (517)

Using the relation between algebraic generators Pmicro and derivatives in the presence of the

vertex operators this implies

part2


F1(eminusiP middotx1v+otimeseiKmiddotxprime2vminusotimeseiKmiddotxprime3vminusotimeseiKmiddotxprime4vminus)=f(x1 x

prime2)f(x1 x

prime3)f(x1 x

prime4) (518)

F1 is determined by F1 up to terms harmonic in x1 We know that x22x23x

24I

(1)(x1 x2 x3 x4)

solves this differential equation so we can identify

x22x23x

24 I(1)(x1 x2 x3 x4)rarr F1(e

minusiP middotx1v+ otimes eiKmiddotxprime2vminus otimes eiKmiddotxprime3vminus otimes eiKmiddotxprime4vminus) (519)

While x23x24I

(1) is an so(4)-equivariant map it is not so(4 2) equivariant even though

the Laplacian in x1 acting on it gives the so(4 2) equivariant map F1 The equivariance

condition under the action of the momentum operator

F1(PmicroeminusiP middotx1v+ otimes eiKmiddotx2vminus otimes eiKmiddotx3vminus otimes eiKmiddotx4vminus)

+F1(eminusiP middotx1v+ otimes PmicroeiKmiddotx2vminus otimes eiKmiddotx3vminus otimes eiKmiddotx4vminus)

+F1(eminusiP middotx1v+ otimes eiKmiddotx2vminus otimes PmicroeiKmiddotx3vminus otimes eiKmiddotx4vminus)

+F1(eminusiP middotx1v+ otimes eiKmiddotx2vminus otimes eiKmiddotx3vminus otimes PmicroeiKmiddotx4vminus) = 0 (520)

ndash 20 ndash

JHEP03(2016)165

implies that (part

partxmicro1+ x22

part

partxmicro2

1

x22+ x23

part

partxmicro3

1

x23+ x24

part

partxmicro4

1

x24

)F1 = 0 (521)

This condition is not satisfied if we identify F1 rarr x22x23x

24I

(1) We can add homogeneous

terms annihilated by 1 to get x22x23x

24(I

(1)+I(2)+I(3)+I(4)) = x22x23x

24I Now equivariance

under Pmicro action of F1 follows from the standard translation invariance of the integral I(part

partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)I = 0 (522)

Similar remarks hold for invariance under the special conformal transformations Kmicro and the

dilatation operator D Hence the quantum equation of motion along with the requirement

of so(4 2) equivariance condition identifies the lift F1 as

F1(eminusiP middotx1v+ otimes eiKmiddotx2vminus otimes eiKmiddotx3vminus otimes eiKmiddotx4vminus) = x22x

23x

24I(x1 x2 x3 x4) =

x22x23x

24

x213x224

Φ(s t)

(523)

By inversion a similar discussion holds for I(4) and the QEOM for x4

F4 V+ otimes V+ otimes V+ otimes Vminus rarr C (524)

with

F4(eminusiP middotx1v+ otimes eminusiP middotx2v+ otimes eminusiP middotx3v+ otimes eiKmiddotxprime4KmicroKmicrov

minus)=xprime21 x

prime22 xprime23

(xprime1minusxprime3)2(xprime2minusxprime3)2(xprime3minusxprime4)2(525)

The x21x22x

23I

(4) integral is an so(4) invariant lift of F4

F4(eminusiP middotx1v+ otimes eminusiP middotx2v+ otimes eminusiP middotx3v+ otimes eiKmiddotxprime4vminus)rarr xprime21 x

prime22 xprime23 I(4)(x1 x2 x3 x4) (526)

The so(4 2) equivariant lift is again given by adding homogeneous terms

F4(eminusiP middotx1v+ otimes eminusiP middotx2v+ otimes eminusiP middotx3v+ otimes eiKmiddotxprime4vminus) = xprime21 x

prime22 xprime23 I(x1 x2 x3 x4) (527)

53 QEOM and twisted equivariant map

In the above discussion the solution I(1) to the QEOM is not logarithmic Logarithmic

contributions are added to ensure so(4 2) equivariance of the lift from V(p2)+ to V+ It is

interesting to see how things are modified when we consider the case of I(2) which is a

logarithmic solution to the quantum equation of motion It is instructive to consider the

free field correlator

〈φ(x1)φ3(x2)φ(x3)φ(x4)〉 (528)

In the range |x1| lt |x2| lt |x3| lt |x4| relevant to I(2) this free-field correlator with

the correct series expansion is constructed by taking φminus(xprime2)otimes φ+(x2)otimes φ+(x2) at x2 and

applying invariant pairings with φ+(x1) and φminus(xprime3)otimesφminus(xprime4) In the free field CFT4TFT2

ndash 21 ndash

JHEP03(2016)165

construction we used φ = φ+ + φminus and the composite field φotimes3 involved sums including

φ+ otimes φ+ otimes φ+ and φminus otimes φ+ otimes φ+ For such sums to be so(4 2) covariant we must use a

twisted co-product

There is a family of automorphisms of so(4 2) parametrized by a number λ

αλ(Pmicro) =Kmicro

λαλ(Kmicro) = λPmicro

αλ(Mmicroν) = Mmicroν αλ(D) = minusD (529)

A homomorphism between so(4 2) and so(4 2)otimes4 is given by the twisted co-product

∆λ(La) = αλ(La)otimes 1otimes 1otimes 1 + 1otimes La otimes 1otimes 1 + 1otimes 1otimes La otimes 1 + 1otimes 1otimes 1otimes La (530)

We can write a new version of (515)

F2(eiP middotx1v+ otimes eminusiP middotx2(PmicroPmicrov

+)otimes eiKmiddotxprime3vminus otimes eiKmiddotxprime4vminus)=x22x

23x

24

(x1 minus x2)2(x2 minus x3)2(x2 minus x4)2(531)

where the so(4 2) acts on the tensor product via the above twisted homomorphism with

the choice λ = x22 We will express this as

F2 V prime+ oplus V(p2)+ otimes Vminus otimes Vminus rarr C (532)

The first factor is written as V prime+ because the twist αλ is being applied In the appendix C2

we show that (531) indeed follows from the equivariance with respect to the twisted co-

product As in the discussion of the x1 QEOM above consider lifts of this map to

F2 V prime+ oplus V+ otimes Vminus otimes Vminus rarr C (533)

In this case converting P middot P into a differential operator is quite subtle This is because

there is x2 dependence in the vertex operator being inserted at the second slot but also

x2 dependence in the twist which determines the map F2 We will leave the problem of

resolving this subtlety as an important technical exercise for the future

A similar discussion applies to I(3) There is an so(4 2) equivariant map

F3 V+ otimes V+ otimes V (p2)minus otimes V primeminus rarr C (534)

which gives the rhs of the quantum equation of motion for x3 In this case we use a

coproduct twisted on the last factor by the automorphism αxprime23

F3(eminusiP middotx1v+ otimes eminusiP middotx2v+ otimesKmicroKmicroe

iKmiddotxprime3vminus otimes eminusiKmiddotxprime4vminus) =xprime21 x

prime22 xprime23


6 Conclusions and future directions

Much of our discussion of the four-point integral in four dimensions should generalize to

the case of the three-point integral in six dimensions and the six-point integral in three

dimensions when we use the appropriate coordinate space propagators

ndash 22 ndash

JHEP03(2016)165

Figure 2 2-loop conformal integral

61 Towards higher loops

We have focused attention on the case of the 1-loop conformal integral We outline how

some key aspects of the discussion generalizes to 2-loops The 2-loop integral is

I2 =

intd4x5d

4x6xminus256

x215x225x

245x

226x

246x

236

(61)

where x2ij = (xi minus xj)2 The exact answer is known It has a term (log(s))2F2(s t)

The term F2(s t) can be recovered from the HPEM of integration Consider the order

|x1| lt |x2| lt |x3| lt |x4| This term arises from the integration range |x2| lt |x5| lt|x6| lt |x3| The expansion of the function F2(s t) has an interpretation in terms of so(4 2)

equivariant maps indicated by the diagram in figure 2 It is an equivariant map acting on

eminusiP middotx1v+otimeseminusiP middotx2v+otimeseminusiP middotx2v+otimeseiKmiddotxprime3vminusotimeseiKmiddotxprime3vminusotimeseiKmiddotxprime4vminus The map is constructed

by composing two projectors one for each integration variable There is an invariant map

pairing two of their indices corresponding to the internal line The projectors are the same

ones we encountered in the 1-loop discussion V+otimesV+ rarr V++ rarr V+otimesV+ It is also possible

to modify the diagram by attaching two external legs to each of x2 x4 respectively (see

figure 3) In that case the x2 x4 become integrated internal vertices The resulting integral

has a fourth power of log which can be recovered from the HPEM method The coefficient

of this logarithmic term is interpreted in terms of a composition of four of the projectors

one for each internal vertex and is closely related to the coefficient of the log-squared in

I2 We leave a more careful exposition of the 2-loop and higher loop cases to a forthcoming

paper but the above statements should be fairly plausible to the attentive reader based

on the discussion in this paper so far

There are two elements to our paper We have developed a group theoretic interpreta-

tion of the coefficient of the log (in sections 3 and 4) Secondly we have developed a group

theoretic interpretation of the complete integral (in section 5) This second goal is more

ambitious and requires properly taking into account indecomposable representations We

expect similar stories to hold at higher loops

ndash 23 ndash

JHEP03(2016)165

Figure 3 Closely-related-integral

62 TFT2 and renormalization

The original motivation for this work was to extend free CFT4TFT2 to interacting the-

ories The discussion in this paper developing the relation between Feynman integrals

and equivariant maps gives some useful clues in this direction for the case of perturbative

interacting CFTs Concrete cases to consider are N = 4 SYM and the Wilson-Fischer fixed

point The connection between quantum equations of motion and indecomposable repre-

sentations we have described should play a role In the free CFT4TFT2 we worked with

a state spaceoplusSymn(V+oplusVminus) where V+ is the irreducible representation obtained from

V+ by quotienting out the V(p2)+ For the interacting CFT4 the state space should involve

tensor products involving V+ and Vminus There will be a coupling dependent quotient given

by the quantum equations of motion Once the correlators are computed at a given order in

the perturbation expansion we know that there is a renormalized formulation where these

correlators are reproduced from local operators having dimensions shifted away from their

values in the free limit A TFT2 formulation of perturbative CFT will presumably incorpo-

rate this renormalization in a sequence of TFT2s one for each order of perturbation theory

such that the correlators computed at any stage of the sequence agree with each other This

will embed the renormalization for CFT4s in a TFT2 set-up the benefit would be to keep

as much as possible the conformal equivariance properties manifest in the process

63 Conformal blocks and CFT4TFT2

The existence of a TFT2 approach to CFT4 is made plausible by several known facts

about CFT4 CFT4 (and in fact more generally CFT in any dimension) has a distinctively

algebraic flavour By the operator-state correspondence local operators correspond to

definite representations of so(4 2) The spectrum of dimensions in the CFT4 along with

the structure constants of the OPE determine the conformal field theory The description

of conformal blocks which exploits Casimirs in a central way has a distinct similarity

to projectors in representation theory (see for example [31ndash33] and more recently in a

superconformal setting [52]) While these facts strongly suggest the existence of a TFT2

formulation the latter is not a trivial consequence For example to understand from a

purely representation theoretic point of view (as required in a TFT2 which by definition is

about equivariant maps) the fact that the OPE of φ2 with φ2 in the free theory contains

ndash 24 ndash

JHEP03(2016)165

φ2 [13] exploits in an important way the representation in TFT2 of the quantum field as

a linear combination involving both V+ and Vminus The ordinary tensor product of V++ otimesV++ does not contain V++ An interesting project in the CFT4TFT2 programme is

to understand in terms of equivariant maps examples in perturbative CFT4 of 4-point

functions where factorization involves analogous OPEs with both the positive dimension

representations and their negative dimension duals playing a role

In special cases of OPEs of the type φ2 otimes φ2 rarr φ4 where the total number of fields

in the intermediate operator is the same as in the external operators which have a direct

analog in the tensor products of positive representations without requiring negative energy

representations in a crucial way there should be a close relation between discussions of

conformal blocks in the physics literature [31ndash33] and the equivariant map interpretation

of integrals developed in [29]

64 HPEM and the interactionintertwiner connection

The HPEM has been an important tool in our discussion of conformal integrals It has

enabled us to make the connection between so(4 2) equivariant maps involving indecom-

posable representations and the quantum equations of motion This connection links a

subtle aspect of representation theory with a consequence of the collision of interaction

point with external vertex a deep and generic property of interactions in quantum field

theory Using the HPEM the full integral was decomposed as a sum of terms I(a) each

involving the collision of the interaction point with one of the external legs and each asso-

ciated with one of the QEOM It will be very interesting to develop this physical picture

for more general Feynman integrals (not necessarily conformal) uncovering the interplay

between the collision of interaction points quantum equations of motion and equivariance

As an example of a simple non-conformal integral consider in four dimensions an

integration in coordinate space of an n-point scalar interaction (with n 6= 4) To interpret

in terms of so(4 2) equivariance we would need to combine the scaling in spacetime with a

scaling of the coupling constant In other words the equivariance would involve a ldquotwistedrdquo

so(4 2)prime which combines the space-time so(4 2) with an so(2) scaling the coupling constant

Twistings which combine space-time symmetries with other symmetries are known to be

useful in topological field theories The idea of employing a scaling of the coupling constant

to arrive at a generalized conformal symmetry was developed in [53]

Acknowledgments

SR is supported by STFC Grant STJ0004691 String Theory Gauge Theory and Duality

RdMK is supported by the South African Research Chairs Initiative of the Department

of Science and Technology and the National Research Foundation We thank Gabriele

Travaglini for very useful discussions at various stages of the project We also thank Andi

Brandhuber Paul Heslop and Donovan Young for useful discussions SR thanks the Simons

summer workshop for hospitality at the Simons Centre for Geometry and Physics and the

Corfu workshop on non-commutative field theory and gravity for hospitality while part of

this work was done

ndash 25 ndash

JHEP03(2016)165

A Basic formulae for HPEM (harmonic polynomial expansion method)

This appendix summarizes the formulae used in the harmonic expansion method for inte-

grals This is also called the Gegenbauer Polynomial expansion technique

so(4) harmonics notation

We will expand the propagators using so(4) spherical harmonics Y lm(x) x is a 4-vector xi

in Euclidean space The positive integer l specifies a symmetric traceless so(4) tensor with

rank l We will work with normalization

1

2π2

intdS3Y l1

m1(x)Y l2

m2(x) = gm1m2

δl1l2

(l1 + 1)(A1)

where dS3 is the standard metric on the unit sphere We could work with more general

bases where the g factor depends on l1 but we wonrsquot use this freedom One convenient

basis is a real orthogonal basis for which

gm1m2 = δm1m2

Y m1l (x) = gm1m2Y l

m2(x) = Y l

m1(x)

Y ml (x) = Y l

m(x) (A2)

Another convenient basis uses the isomorphism so(4) = su(2)times su(2) A rank l symmetric

tensor specifies a representation of spins ( l2 l2) The so(4) state label m is equivalent to a

pair of su(2) state labels (mLmR) each ranging from minus l2 to l

2 If we work with a basis

which diagonalize the JL3 JR3 then the state label is equivalent to a pair of su(2) labels

(mLmR)

gm1m2 = δm1minusm2

Y ml (x) = Y l

minusm(x) (A3)

Using the explicit generators JLi JRi given in [13] the charges (JL3 J

R3 ) for the basic vari-

ables are

z = x1 + ix2 harr(

1

2

1

2

)z = x1 minus ix2 harr

(minus1

2minus1

2

)w = x3 + ix4 harr

(1

2minus1

2

)w = x3 minus ix4 harr

(minus1

21

2

)(A4)

We find that

Y l(l2l2) = (x1 + ix2)

l = zl (A5)

has normalization

1

2π2

intdS3(Y l

( l2 l2))lowast Y l

( l2 l2)

=1

(l + 1)(A6)

The remaining spherical harmonics are easily generated using the su(2) lowering operators

ndash 26 ndash

JHEP03(2016)165

Expansion of the exponential vertex operator The expansion of the exponential

in terms of spherical harmonics is

eminusiP middotxv+ =sumlm

(minusi)l

2llY lm(x)Y m

l (P )v+ (A7)

The invariant pairing η V+ otimes Vminus rarr C described in terms of harmonics in PK is

η(Y l1m1

(P )v+ Y m2l2

(K)vminus)

= δl1l2δm1m222l1(l1)2 (A8)

The invariant pairing in terms of spherical harmonics in x ie η H+ otimesHminus rarr C is

η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)= δl1l2δm1m2 (A9)

This can be written in terms of integration on the unit 3-sphere [29]

η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)=

1

2π2

intS3

Y m2l2

(xprime)

xprime2(x middot partx + 1)Y l1

m1(x)) (A10)

There are so(4 2) equivariant maps

f+ V+ rarr Hminusfminus Vminus rarr H+ (A11)

which can be read off from the expansion of the exponential

f+ Y lm(P )v+ rarr 2ll(i)l

Y lm(xprime)

xprime2

fminus Y lm(K)vminus rarr 2ll(minusi)lY l

m(x) (A12)

Combining the two maps f+ otimes fminus = f gives a map f V+ otimes Vminus rarr Hminus otimes H+ We have

η = f η

Expansion of the 2-point function

1

|~xminus ~y|2=infinsuml=0

summ

1

y2Y lm(x)Y m

l (yprime) if |y| gt |x|

=infinsuml=0

summ

1

x2Y lm(xprime)Y m

l (y) if |x| gt |y| (A13)

Using the addition theorem for spherical harmonics the rhs can be written in terms of

Gegenbauer polynomials of x middot y This is a well-known way of doing complicated inte-

grals [42] We will not be writing expansions in terms of Gegenbauer polynomials since

our main purpose is to keep all four so(4 2)rsquos associated with the external legs manifest

rather than finding an efficient way to do the integrals

ndash 27 ndash

JHEP03(2016)165

Action of the quadratic so(4 2) Casimir

C2 = minus1

2LMNL

MN = D2 minus 4D + ~P middot ~K +1

2MpqMqp (A14)

We will use the differential operator representation of the generators to compute the value

of C2 when acting on a product of a function of r and a spherical harmonic As usual

we use vertex operators to obtain the differential opetator corresponding to a particular

generator For example

~P middot ~KeminusiP middotxv+ = Pp

(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)eminusiP middotxv+

=

(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)ipart

partxpeminusiP middotxv+ (A15)

To obtain the first equality for example commute Kmicro past the vertex operator eminusiP middotx using

the so(4 2) algebra and then use the fact that Kmicrov+ = 0 Finally express the result as a

differential operator acting on the vertex operator Similarly we find

Mpq = xppart

partxqminus xq

part

partxp(A16)

so that when acting on a power of r times a spherical harmonic we have

1

2MpqMqp(r

AY lm(x)) =

rA

2MpqMqpY

lm(x) = l(l + 2) rAY l

m(x) (A17)

We also have (D = x middot partpartx + 1 is shifted by 1 to account for the dimension of v+)

(D2 minus 4D) rAY (x) = [(A+ l minus 2)(A+ l)minus 3] rAY lm(x) (A18)

Finally consider ~P middot ~K

~P middot ~KrAY lm(x) = (A(A+ 2) + 2Al minus 2(A+ l)2) rAY l

m(x) (A19)

Thus we have

C2 rAY l

m(x) = minus3 rAY lm(x) (A20)

A very similar argument shows that

C2 log(r)Y lm(x) = minus3 log(r)Y l

m(x) (A21)

B Expansion of projector using the exact answer

This section extends the discussion of section 22 by providing some of the details behind

the expansion We want to study the limit s rarr 0 t rarr 1 ie u rarr 0 where t = 1 + u In

this limit λrarr 0 and ρrarrinfin The coefficient of the log s is

F1 =1

λlog

((1 + ρs)


)(B1)

ndash 28 ndash

JHEP03(2016)165

Using Mathematica we expand in s and simplify the function of u appearing at each order

of s to obtain

F1 =log(1+u)

u+

s

u3(2uminus (2+u) log(1+u)) +

s2

u5(3u(2+u)minus 6 + 6u+ u2) log(1+u)

)+s3

3u7(u(60 + 60u+ 11u2)minus 3(20 + 30u+ 12u2 + u3) log(1 + u)

)(B2)

After subsequently expanding in powers of u we have

F1 =

(minus1 +

u

2minus u2

3+u3

4minus middot middot middot

)+ s

(minus1

6+u

6minus 3u2

20+ middot middot middot

)+ s2

(minus1

30+

u

20minus 2u2


)+s3

(minus1

140+

u

70minus 5u2


)+ middot middot middot (B3)

The term at sk is

sk

u2k+1(uPk(u)minusQk(u) log(1 + u)) (B4)

where Pk(u) is a polynomial of degree k minus 1 and Qk(u) is a polynomial of degree k both

with positive coefficients The polynomials have the property that the expansion at sk is

regular at u = 0 In other words

uPk(u)minusQk(u) log(1 + u) (B5)

only contains powers un with ngt2k This gives 2k+1 equations constraining the (2k+1)

unknown coefficients in Pk and Qk These equations do not determine the overall normal-

ization of the two polynomials This is determined by observing that the leading coefficient

in Pk(u) is

(2k)

(k)2(B6)

which is (k + 1) times the krsquoth Catalan number Writing

Qk(u) =ksuml=0

blul (B7)

we find the linear system of equations

ksuml=0

bl(minus1)l

(iminus l)= 0 (B8)

as i ranges from k + 1 to 2k These come from the requirement that the coefficient of

xi vanishes in (B5) These equations allow us to solve b1 b2 middot middot middot bk in terms of b0 For

example when k = 5 we have

b1 =5b02 b2 =

20b09 b3 =

5b06 b4 =

5b042 b5 =

b0252

(B9)

ndash 29 ndash

JHEP03(2016)165

Interestingly the Catalan number 252 comes from solving this system of linear equations

To solve (B8) for any k define I = iminus k where the range of I is 1 le I le k and we have

ksuml=0

MIlbl = 0 (B10)

where

MIl =(minus1)l

(I + k minus l)(B11)

The index l ranges over k + 1 values We can rewrite (B10) as

MI0b0 = minusksuml=1

MIlbl (B12)

Define Mab = Mab with a b = 1 k Using Mathematica to study a few examples we

have verified that M is invertible so that we can write

bl =sumI

Mminus1JI MI0b0 (B13)

For a specific choice of k it is easy to generate the inverses of M in Mathematica and

hence to generate the bIb0 For k = 2 7 we find

bIb0 = 1 16bIb0 = 32 35 120bIb0 = 2 97 27 170

bIb0 =

5

2

20

95

6

5

42

1

252

bIb0 =

3

75

22

20

11

5

11

1

22

1

924

bIb0 =

7

2

63

13175

52

175

143

63

286

7

429

1

3432

(B14)

Using these numerical results from Mathematica and the Online Encyclopaedia of Integer

Sequences [54] we find

b1(k)b0 = k2

bk(k)b0 =k2

(2k)

bkminus1(k)b0 =k2

(2k)k(k + 1)

bkminus2(k)b0 =k2

(2k)

(k minus 1)k(k + 1)(k + 2)

4

bkminus3(k)b0 =k2

(2k)

(k

3

)(k + 3

3

)bkminusm(k)b0 =

k2

(2k)

(k

m

)(k +m

m

)(B15)

ndash 30 ndash

JHEP03(2016)165

This gives the general formula for the b-coefficients We also know b0 = (2k)k2

so that

we have

b1(k) = k2(2k)

k2

bk(k) = 1

bkminus1(k) = k(k + 1)

bkminus2(k) =(k minus 1)k(k + 1)(k + 2)

4

bkminus3(k) =

(k

3

)(k + 3

3

)bkminusm(k) =

(k

m

)(k +m

m

)(B16)

Lets us now consider the polynomial Pk(u)

Pk(u) =

ksumi=1

aiuiminus1 (B17)

By looking at powers ui in (B5) for 1 le i le k we obtain the linear equations

ai minusiminus1suml=0

bl(minus1)iminusl

(iminus l)= 0 (B18)

This gives the ai as sums of binomial coefficients using the formula for b above We can

again numerically work out the a-coefficients for small values of k and then read off the

analytic formulas from the patterns we find For the first few values of k we find

k = 2 ~a = a1 ak = 6 3

k = 3 ~a =1

3times 60 60 11

k = 4 ~a =5

6times 84 126 52 5

k = 5 ~a =1

30times 7560 15120 9870 2310 137

k = 6 ~a =7

10times 13720 3300 2960 1140 174 7 (B19)

which leads to

a1(k) =(2k)

(k)2

ak(k) = 2h(k) = 2

ksuml=1

1

l(B20)

h(k) is the k-th harmonic number Thus the ai(k) interpolate between Catalan numbers

and harmonic numbers In this way the Catalan numbers along with the form (B4) has

determined all the coefficients in the double Taylor expansion around s u = 0

ndash 31 ndash

JHEP03(2016)165

B1 A summation formula for products of su(2) Clebschs from Feynman inte-

grals

Our discussion implies a summation formula for products of su(2)-Clebsh Gordan coeffi-

cients since these coefficients appear in the multiplication of spherical harmonics which

enter the definition of the so(4 2) equivariant map M Indeed equating the result for the

coefficient of the log obtained from HPEM to the coefficient of the same log appearing in

the exact result we have

infinsumli=0

summi

infinsumt=0

sump

δl1+l2l3+l4Yl1minusm1

(x1)Yl2minusm2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)timesC l1l2tm1m2pC

l3l4tminusm3minusm4minusp

t+ 1

=x23x

24

x213x224

1

λlog

((1 + ρs)


)(B21)

Noting that the structure constants of the multiplication of so(4)-covariant harmonics on

S3 can be written in terms of Clebsch-Gordan coefficients of su(2)

C l1 l2 lm1m2 m1+m2

=

langl12

l22

mL12

mL22

∣∣∣∣∣ l2

mL1 +mL2

2

ranglangl12

l22

mR12

mR22

∣∣∣∣∣ l2

mR1 +mR22

rangwe see that (B21) is a highly nontrivial sum rule for su(2)-Clebsh Gordan coefficients

To check this sum rule it is useful to use the basis which diagonalizes the (JL3 JR3 )

sub-algebra of su(2)times su(2) described in (A4) Using this basis we can easily determine

the coefficients of monomials of a specific form appearing on both side of (B21) The

simplest monomial arises from the terms in the sum with t = 0 and

Yn12

n12n12

(x1)Yn22

n22n22

(x2)Yn32minusn32minusn32


(x4) = zn11 zn2

2 zn33 zn4

4 (B22)

In this extremal case the Clebsch Gordan coefficients are 1 so that the monomial

zn11 zn2

2 zn33 zn4

4 has coefficient 1n1+n2+1 To recover this coefficient from the rhs of (B21)

note that

(xi minus xj)2 = 1minus 2xi middot xj + x2ix2j

= 1minus zizj minus zizj minus wiwj minus wiwj + (zizi + wiwi)(zj zj + wjwj) (B23)

Inserting this into the rhs of (B21) and expanding as described at the start of this

appendix we can obtain the coefficient of any given monomial In particular we verify

that zn11 zn2

2 zn33 zn4

4 has coefficient 1n1+n2+1 Next consider

Yn12

n12n1minus2

2

(x1)Yn22

n2minus22

n22

(x2)Yn32

minusn32minusn3minus2

2

(x3)Yn42

minusn4minus22

minusn42

(x4)

=radicn1n2n3n4z

n1minus11 w1z

n2minus12 w2z

n3minus13 w3z

n4minus14 w4 (B24)

which involves terms in the sum with t = 0 and t = 1 We need two Clebsch Gordan

coefficients

Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B25)

Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B26)

ndash 32 ndash

JHEP03(2016)165

which easily follow from the following su(2) coefficients

Cn12n22n1+n2

2n12n2minus2

2n1+n2minus2

2

=

radicn2

n1 + n2(B27)

Cn12n22n1+n2minus2

2n12n2minus2

2n1+n2minus2

2

= minusradic

n1n1 + n2

(B28)

Explicit formulae for the su(2) Clebsch Gordan coefficients are available in [55] Thus the

coefficient of zn1minus11 w1z

n2minus12 w2z

n3minus13 w3z

n4minus14 w4 is

radicn1n2n3n4

[Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2 + 1

+Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2

]=

2n1n2n3n4(n1 + n2 minus 1)(n1 + n2)(n1 + n2 + 1)

(B29)

This again agrees with the coefficient obtained by expanding the rhs of (B21)

C Equivariant maps related to quantum equations of motion

This section gives the explicit evaluation of the maps F1 and F2 which are introduced in

section 5

C1 Quantum equation of motion for x1

When we evaluate the map F1 with the exponential states inserted we get an expres-

sion which has a well defined expansion at small x1 xprime2 xprime3 xprime4 Set h(x1 x

prime2 xprime3 xprime4) =

F1(eminusiP middotx1v+++ otimes eiKmiddotxprime2vminus otimes eiKmiddotxprime3vminus otimes eiKmiddotxprime4vminus) Then

parth

partxmicro1= minusiF1(Pmicroe

minusiP middotx1v+++ otimes eiKmiddotxprime2vminus otimes eiKmiddotxprime3vminus otimes eiKmiddotxprime4vminus)

= iF1(eminusiP middotx1v+++ otimes PmicroeiKmiddotx

prime2vminus otimes eiKmiddotxprime3vminus otimes eiKmiddotxprime4vminus)

+iF1(eminusiP middotx1v+++ otimes eiKmiddotxprime2vminus otimes PmicroeiKmiddotx

prime3vminus otimes eiKmiddotxprime4vminus)

+iF1(eminusiP middotx1v+++ otimes eiKmiddotxprime2vminus otimes eiKmiddotxprime3vminus otimes PmicroeiKmiddotx

prime4vminus) (C1)

where the last line follows from the so(4 2) invariance of F1 Now

PmicroeiKmiddotxprime3vminus =

(2ixprime3microx

prime3 middot

part

partxprime3minus ixprime23

part

partxprime3micro+ 2ixprime3micro

)eiKmiddotx

prime3vminus

= ix23part

partxmicro3

1

x23eiKmiddotx

prime3vminus (C2)

so that (C1) becomes

parth

partxmicro1= minusx22

part

partxmicro2

(1

x22h

)minus x23

part

partxmicro3

(1

x23h

)minus x24

part

partxmicro4

(1

x24h

)(C3)

ndash 33 ndash

JHEP03(2016)165

which can be written as

x22x23x

24

(part

partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)[h

x22x23x

24

]= 0 (C4)

This proves the map is only a function of the differences xmicroi minus xmicroj Next consider(

x1micropart

partx1νminus x1ν

part

partx1micro

)h= F1(Mmicroνe


=minusF1(eminusiP middotx1v+++ otimesMmicroνe

iKmiddotxprime2vminus otimes eiKmiddotxprime3vminus otimes eiKmiddotxprime4vminus)

minusF1(eminusiP middotx1v+++ otimes eiKmiddotxprime2vminus otimesMmicroνe

iKmiddotxprime3vminus otimes eiKmiddotxprime4vminus)

minusF1(eminusiP middotx1v+++ otimes eiKmiddotxprime2vminus otimes eiKmiddotxprime3vminus otimesMmicroνe

iKmiddotxprime4vminus) (C5)

which after a little algebra can be written as

x22x23x

24

(x1micro

part

partx1νminus x1ν

part

partx1micro+ x2micro

part

partx2νminus x2ν

part


part

partx3νminus x3ν

part

partx3micro

+x4micropart

partx4νminus x4ν

part

partx4micro

)[h

x22x23x

24

]= 0 (C6)

This proves the map is only a function of the magnitudes of the differences |xmicroi minusxmicroj | Next

consider(x1 middot

part

partx1+ 3

)h = F1(e


= minusF1(eminusiP middotx1v+++ otimesDeiKmiddotxprime2vminus otimes eiKmiddotxprime3vminus otimes eiKmiddotxprime4vminus)

minusF1(eminusiP middotx1v+++ otimes eiKmiddotxprime2vminus otimesDeiKmiddotxprime3vminus otimes eiKmiddotxprime4vminus)

minusF1(eminusiP middotx1v+++ otimes eiKmiddotxprime2vminus otimes eiKmiddotxprime3vminus otimesDeiKmiddotxprime4vminus) (C7)

This can be rewritten as

x22x23x

24

(x1 middot

part

partx1+ x2 middot

part

partx2+ x3 middot

part

partx3+ x4 middot

part

partx4+ 12

)[h

x22x23x

24

]= 0 (C8)

This tells us the degree of the dependence on |xmicroi minus xmicroj | Thus at this stage we know that

h

x22x23x

24

=A

|x1 minus x2|2α|x1 minus x3|2β |x1 minus x4|2γ |x2 minus x3|2δ|x2 minus x4|2η|x3 minus x4|2τ(C9)

and α + β + γ + δ + η + τ = 6 Thus the map has now been reduced to determining 7

numbers To determine these numbers start by noting that at x1 = 0 and xprime2 = xprime3 = xprime4 = 0

we have

h(0 0 0 0) = F1(v+++ otimes vminus otimes vminus otimes vminus) = 1 (C10)

ie h has no singularities and takes the constant value 1 This implies that δ = η = τ = 0

α = β = γ = 1 and A = 1 which proves (513)

ndash 34 ndash

JHEP03(2016)165


Using invariance of the map and the so(4 2) algebrawe can easily prove (531) Towards

this end recall some results which follow from the invariant pairing

η(eminusiP middotx1v+ eiKmiddotxprime2vminus) =

infinsumnm=0

im(minus1)n

nmxα11 middot middot middotx

αn1 xprimeβ12 middot middot middotx

primeβm2 Tα1middotmiddotmiddotαnβ1middotmiddotmiddotβm (C11)

where

Tα1middotmiddotmiddotαnβ1middotmiddotmiddotβm = η(Pα1 middot middot middotPαnv+Kβ1 middot middot middotKβmvminus)

= (minus1)mη(v+ Pαn middot middot middotPα1Kβ1 middot middot middotKβmvminus)

= (minus1)mη(Kβm middot middot middotKβ1Pα1 middot middot middotPαnv+ vminus) (C12)

and the last two lines above follow from the so(4 2) invariance of the pairing Now setting

Kβm middot middot middotKβ1Pα1 middot middot middotPαnv+ = δnmtα1middotmiddotmiddotαnβ1middotmiddotmiddotβmv+ (C13)

Pαn middot middot middotPα1Kβ1 middot middot middotKβmvminus = δnmtα1middotmiddotmiddotαnβ1middotmiddotmiddotβmv

minus (C14)

we find

Tα1middotmiddotmiddotαnβ1middotmiddotmiddotβm = tα1middotmiddotmiddotαnβ1middotmiddotmiddotβm(minus1)m (C15)

and

infinsumn=0

(minus1)n



primeβn2 tα1middotmiddotmiddotαnβ1middotmiddotmiddotβn =

1

1minus 2xprime2 middot x1 + x21xprime22

(C16)

We will make use of (C13) (C14) and (C16) below Consider the complete expansion

F2(eiP middotx1v+ otimes eminusiP middotx2v+++ otimes eiKmiddotxprime3vminus otimes eiKmiddotxprime4vminus) (C17)

=infinsum

n1n3n4=0

1

n1n3n4F2((iP middot x1)n1v+ otimes eminusiP middotx2v+++ otimes (iK middot xprime3)n3vminus otimes (iK middot xprime4)n4vminus)

Expand the remaining exponential and use equivariance of the map to transfer the P middot x2factors into the other three slots Due to the twisting when we move P middot x2 into the first

slot we get

α(P middot x2) =1

x22K middot x2 = K middot xprime2 (C18)

ndash 35 ndash

JHEP03(2016)165

For a given term in the sum (ie a given n1 n3 n3) only a specific power of P middotx2 from the

expansion of the exponential in slot 2 will contribute Keeping only this power we have

F2(eiP middotx1v+ otimes eminusiP middotx2v+++ otimes eiKmiddotxprime3vminus otimes eiKmiddotxprime4vminus)

=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)

timesF2((iP middot x1)n1v+ otimes (minusiP middot x2)n1+n3+n4v+++ otimes (iK middot xprime3)n3vminus otimes (iK middot xprime4)n4vminus)

=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)

(n1 + n3 + n4)

n1n3n4(minus1)n1+n3+n4

timesF2((α(P middot x2))n1(P middot x1)n1v+otimesv+++otimes(P middot x2)n3(K middot xprime3)n3vminusotimes(P middot x2)n4(K middot xprime4)n4vminus)

=infinsumn1

(minus1)n1

(n1)2xα11 middot middot middotx


primeβn12 tα1middotmiddotmiddotαn1β1middotmiddotmiddotβn1

timesinfinsumn3

(minus1)n3

(n1)2xγ12 middot middot middotx

γn32 xprimeδ13 middot middot middotx

primeδn33 tγ1middotmiddotmiddotγn3δ1middotmiddotmiddotδn3

timesinfinsumn4

(minus1)n4

(n1)2xmicro12 middot middot middotx

micron42 xprimeν14 middot middot middotx

primeνn44 tmicro1middotmiddotmiddotmicron4ν1middotmiddotmiddotνn4F2(v

+ otimes v+++ otimes vminus otimes vminus)

=1

1minus 2x1 middot xprime2 + x21xprime22

1


1


=x22x

33x

24

(x1 minus x2)2(x2 minus x3)2(x2 minus x4)2(C19)

which proves the result (531)

Open Access This article is distributed under the terms of the Creative Commons

Attribution License (CC-BY 40) which permits any use distribution and reproduction in

any medium provided the original author(s) and source are credited

References

[1] R de Mello Koch and S Ramgoolam Strings from Feynman Graph counting without

large-N Phys Rev D 85 (2012) 026007 [arXiv11104858] [INSPIRE]

[2] R de Mello Koch S Ramgoolam and C Wen On the refined counting of graphs on surfaces

Nucl Phys B 870 (2013) 530 [arXiv12090334] [INSPIRE]

[3] J Pasukonis and S Ramgoolam Quivers as Calculators Counting Correlators and

Riemann Surfaces JHEP 04 (2013) 094 [arXiv13011980] [INSPIRE]

[4] Y Kimura Multi-matrix models and Noncommutative Frobenius algebras obtained from

symmetric groups and Brauer algebras Commun Math Phys 337 (2015) 1

[arXiv14036572] [INSPIRE]

[5] V Jejjala S Ramgoolam and D Rodriguez-Gomez Toric CFTs Permutation Triples and

Belyi Pairs JHEP 03 (2011) 065 [arXiv10122351] [INSPIRE]

[6] TW Brown Complex matrix model duality Phys Rev D 83 (2011) 085002


ndash 36 ndash

JHEP03(2016)165

[7] S Corley A Jevicki and S Ramgoolam Exact correlators of giant gravitons from dual

N = 4 SYM theory Adv Theor Math Phys 5 (2002) 809 [hep-th0111222] [INSPIRE]

[8] R de Mello Koch M Dessein D Giataganas and C Mathwin Giant Graviton Oscillators

JHEP 10 (2011) 009 [arXiv11082761] [INSPIRE]

[9] R de Mello Koch and S Ramgoolam A double coset ansatz for integrability in AdSCFT


[10] J Ben Geloun and S Ramgoolam Counting Tensor Model Observables and Branched Covers

of the 2-Sphere arXiv13076490 [INSPIRE]

[11] JA Minahan and K Zarembo The Bethe ansatz for N = 4 super Yang-Mills JHEP 03

(2003) 013 [hep-th0212208] [INSPIRE]

[12] N Beisert C Kristjansen and M Staudacher The Dilatation operator of conformal N = 4

super Yang-Mills theory Nucl Phys B 664 (2003) 131 [hep-th0303060] [INSPIRE]

[13] R de Mello Koch and S Ramgoolam CFT4 as so(4 2)-invariant TFT2 Nucl Phys B 890

(2014) 302 [arXiv14036646] [INSPIRE]

[14] M Atiyah Topological quantum field theory Publ Math IHES 68 (1988) 175

[15] GW Moore and G Segal D-branes and k-theory in 2D topological field theory

hep-th0609042 [INSPIRE]

[16] JM Maldacena The large-N limit of superconformal field theories and supergravity Int J

Theor Phys 38 (1999) 1113 [hep-th9711200] [INSPIRE]

[17] SS Gubser IR Klebanov and AM Polyakov Gauge theory correlators from noncritical

string theory Phys Lett B 428 (1998) 105 [hep-th9802109] [INSPIRE]

[18] E Witten Anti-de Sitter space and holography Adv Theor Math Phys 2 (1998) 253

[hep-th9802150] [INSPIRE]

[19] Y Kazama S Komatsu and T Nishimura Novel construction and the monodromy relation

for three-point functions at weak coupling JHEP 01 (2015) 095 [Erratum ibid 1508 (2015)

145] [arXiv14108533] [INSPIRE]

[20] Y Jiang I Kostov A Petrovskii and D Serban String Bits and the Spin Vertex Nucl

Phys B 897 (2015) 374 [arXiv14108860] [INSPIRE]

[21] B Basso S Komatsu and P Vieira Structure Constants and Integrable Bootstrap in Planar

N = 4 SYM Theory arXiv150506745 [INSPIRE]

[22] LF Alday JR David E Gava and KS Narain Towards a string bit formulation of N = 4

super Yang-Mills JHEP 04 (2006) 014 [hep-th0510264] [INSPIRE]

[23] L Freidel RG Leigh and D Minic Quantum Gravity Dynamical Phase Space and String

Theory Int J Mod Phys D 23 (2014) 1442006 [arXiv14053949] [INSPIRE]

[24] L Freidel RG Leigh and D Minic Metastring Theory and Modular Space-time JHEP 06

(2015) 006 [arXiv150208005] [INSPIRE]

[25] MS Costa J Penedones D Poland and S Rychkov Spinning Conformal Correlators


[26] LJ Dixon A brief introduction to modern amplitude methods arXiv13105353 [INSPIRE]

[27] JM Drummond J Henn VA Smirnov and E Sokatchev Magic identities for conformal

four-point integrals JHEP 01 (2007) 064 [hep-th0607160] [INSPIRE]

ndash 37 ndash

JHEP03(2016)165

[28] JM Drummond GP Korchemsky and E Sokatchev Conformal properties of four-gluon

planar amplitudes and Wilson loops Nucl Phys B 795 (2008) 385 [arXiv07070243]

[INSPIRE]

[29] I Frenkel and M Libine Quaternionic Analysis Representation Theory and Physics

arXiv07112699 [INSPIRE]

[30] N Aizawa and VK Dobrev Intertwining Operator Realization of anti de Sitter Holography

Rept Math Phys 75 (2015) 179 [arXiv14062129] [INSPIRE]

[31] FA Dolan and H Osborn Implications of N = 1 superconformal symmetry for chiral fields

Nucl Phys B 593 (2001) 599 [hep-th0006098] [INSPIRE]

[32] FA Dolan and H Osborn Conformal four point functions and the operator product

expansion Nucl Phys B 599 (2001) 459 [hep-th0011040] [INSPIRE]

[33] D Simmons-Duffin Projectors Shadows and Conformal Blocks JHEP 04 (2014) 146


[34] NI Usyukina and AI Davydychev Exact results for three and four point ladder diagrams

with an arbitrary number of rungs Phys Lett B 305 (1993) 136 [INSPIRE]

[35] NI Usyukina and AI Davydychev Some exact results for two loop diagrams with three and

four external lines Phys Atom Nucl 56 (1993) 1553 [Yad Fiz 56N11 (1993) 172]

[hep-ph9307327] [INSPIRE]

[36] S Rychkov and ZM Tan The ε-expansion from conformal field theory J Phys A 48

(2015) 29FT01 [arXiv150500963] [INSPIRE]

[37] FA Dolan and H Osborn On short and semi-short representations for four-dimensional

superconformal symmetry Annals Phys 307 (2003) 41 [hep-th0209056] [INSPIRE]

[38] J Kinney JM Maldacena S Minwalla and S Raju An Index for 4 dimensional super

conformal theories Commun Math Phys 275 (2007) 209 [hep-th0510251] [INSPIRE]

[39] M Bianchi PJ Heslop and F Riccioni More on La Grande Bouffe JHEP 08 (2005) 088


[40] M Libine The Two-Loop Ladder Diagram and Representations of U(2 2) arXiv13095665

[INSPIRE]

[41] M Libine The Conformal Four-Point Integrals Magic Identities and Representations of

U(2 2) arXiv14072507 [INSPIRE]

[42] AV Kotikov The Gegenbauer polynomial technique The Evaluation of a class of Feynman

diagrams Phys Lett B 375 (1996) 240 [hep-ph9512270] [INSPIRE]

[43] R Britto F Cachazo and B Feng Generalized unitarity and one-loop amplitudes in N = 4

super-Yang-Mills Nucl Phys B 725 (2005) 275 [hep-th0412103] [INSPIRE]

[44] F Cachazo Sharpening The Leading Singularity arXiv08031988 [INSPIRE]

[45] S Abreu R Britto C Duhr and E Gardi From multiple unitarity cuts to the coproduct of

Feynman integrals JHEP 10 (2014) 125 [arXiv14013546] [INSPIRE]

[46] MR Gaberdiel Fusion rules and logarithmic representations of a WZW model at fractional

level Nucl Phys B 618 (2001) 407 [hep-th0105046] [INSPIRE]

[47] R Vasseur JL Jacobsen and H Saleur Indecomposability parameters in chiral Logarithmic

Conformal Field Theory Nucl Phys B 851 (2011) 314 [arXiv11033134] [INSPIRE]

ndash 38 ndash

JHEP03(2016)165

[48] A-L Do and M Flohr Towards the construction of Local Logarithmic Conformal Field

Theories Nucl Phys B 802 (2008) 475 [arXiv07101783] [INSPIRE]

[49] FA Dolan and H Osborn Conformal partial waves and the operator product expansion


[50] FA Dolan Character formulae and partition functions in higher dimensional conformal

field theory J Math Phys 47 (2006) 062303 [hep-th0508031] [INSPIRE]

[51] W Heidenreich Tensor Products of Positive Energy Representations of SO(3 2) and

SO(4 2) J Math Phys 22 (1981) 1566 [INSPIRE]

[52] R Doobary and P Heslop Superconformal partial waves in Grassmannian field theories


[53] A Jevicki Y Kazama and T Yoneya Generalized conformal symmetry in D-brane matrix

models Phys Rev D 59 (1999) 066001 [hep-th9810146] [INSPIRE]

[54] The On-Line Encyclopedia of Integer Sequences available at httpoeisorg

[55] httpsenwikipediaorgwikiTable of ClebschE28093Gordan coefficients

ndash 39 ndash

Introduction

Background and motivations

CFT4TFT2 suggests equivariant interpretation of perturbative Feynman integrals

Conformal integral exact answer and an expansion

Indecomposable representations and multiplet recombination

Harmonic expansion method and the logarithmic term

Coefficient of the log term and the projector

Analytic consequences

Quantum equation of motion indecomposable representations and equivariant maps

Quantum equations of motion

QEOM equivariant maps and their lifts

QEOM and twisted equivariant map

Conclusions and future directions

Towards higher loops

TFT2 and renormalization

Conformal blocks and CFT4TFT2

HPEM and the interactionintertwiner connection

Basic formulae for HPEM (harmonic polynomial expansion method)

Expansion of projector using the exact answer

A summation formula for products of su(2) Clebschs from Feynman integrals

Equivariant maps related to quantum equations of motion

Quantum equation of motion for x(1)


JHEP03(2016)165

Contents

1 Introduction 1

2 Background and motivations 5


integrals 5

22 Conformal integral exact answer and an expansion 6

23 Indecomposable representations and multiplet recombination 8

3 Harmonic expansion method and the logarithmic term 9

4 Coefficient of the log term and the projector 13

41 Analytic consequences 17

5 Quantum equation of motion indecomposable representations and equiv-

ariant maps 17

51 Quantum equations of motion 18

52 QEOM equivariant maps and their lifts 19

53 QEOM and twisted equivariant map 21

6 Conclusions and future directions 22

61 Towards higher loops 23

62 TFT2 and renormalization 24

63 Conformal blocks and CFT4TFT2 24

64 HPEM and the interactionintertwiner connection 25

A Basic formulae for HPEM (harmonic polynomial expansion method) 26

B Expansion of projector using the exact answer 28

B1 A summation formula for products of su(2) Clebschs from Feynman integrals 32

C Equivariant maps related to quantum equations of motion 33



1 Introduction

Many aspects of the combinatorics of N = 4 super-Yang-Mills theories have been shown

to be captured by two-dimensional topological field theories (TFT2s) based on permu-

tation groups [1ndash4 6ndash10] Specifically these topological field theories were found in the

ndash 1 ndash

JHEP03(2016)165


























Y lm(P )v+ (13)








Y lm(K)vminus (14)



ndash 2 ndash

JHEP03(2016)165



Φ(x) =1radic2




x2 It is found that

η(Φ(x1)Φ(x2)) =1

(x1 minus x2)2(16)


































ndash 3 ndash

JHEP03(2016)165











































ndash 4 ndash

JHEP03(2016)165















integrals














I(x1 x2 x3 x4) =

intd4x

2π21
















ndash 5 ndash

JHEP03(2016)165







prime4vminus (24)

To this we tensor

































ndash 6 ndash

JHEP03(2016)165


I(x1 x2 x3 x4) =1

2x213x224

Φ(s t) (26)

where

Φ(s t) =1

λ


(t

s

)ln

(1 + ρt

1 + ρs

)+π2

3

)(27)

and

ρ =2



s =x212x

234

x213x224

t =x214x

223

x213x224

(28)



λ sim u+ s

(minus1minus 2

u

)+ s2

(minus 2

u3minus 2

u2

)sim u+ s

(minus1minus 2

u

)ρ sim s2

(2

u5+

1

u4

)+

s


1


s(29)



6minus 1

2log2(minusz) (210)


Φ(s t) =1

λ


(st

)log


(1 + ρs)

)(211)


notation


F0(s u) =1

λ



(1 + ρs)

)F1(s u) = minus 1

λlog


(1 + ρs)









=f(x1 x

prime4)f(x2 x

prime3)


ndash 7 ndash

JHEP03(2016)165

where








in section 4









J3

∣∣∣minus 1

2

rang= minus1

2

∣∣∣minus 1

2

rangJminus

∣∣∣minus 1

2

rang= 0 (214)





JminusJ2+

∣∣∣minus1

2

rang= 0 (215)




(2)

minus 12

The quotient

space Vminus 12V

(2)

minus 12


sequence

0rarr V(2)

minus 12


2V

(2)

minus 12

rarr 0 (216)









ndash 8 ndash

JHEP03(2016)165










(p2)+ rarr 0 (220)
























in section 5


I1 =sumllimi

prod4i=1 Y

limi(x

primei)

|x1|2|x2|2|x3|2|x4|2

int |x1|0


4prodi=1

Y mili

(x) (31)

ndash 9 ndash

JHEP03(2016)165

where d3x = 12π2dS



write it as

Cm1m2m3m4

l1l2l3l4=

suml5m5m6

Cm1m2l5l1l2m5

1

(l5 + 1)Cm3m4l5l3l4m6


(32)

where

Y m1l1

Y m2l2

=suml5m5

Cm1m2l5l1l2m5

Y m5l5

(33)




or equivalently












I1 =sumlimi

prodi Y

limi(x

primei)

|x1|2|x2|2|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

|x1|4+l1+l2+l3+l44 + l1 + l2 + l3 + l4

(38)




ndash 10 ndash

JHEP03(2016)165


region is

I2 =sumlimi

Y l1m1

(x1)prodi=2 Y

limi(x

primei)

|x2|2|x3|2|x4|2

int |x2||x1|


l1(xprime)Y m2

l2(x)Y m3

l3(x)Y m4

l4(x)

=sumlimi

Y l1m1

(x1)prodi=2 Y

limi(x

primei)

|x2|2|x3|2|x4|2

int |x2||x1|


l1l2l3l4

=sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

|x2|2Y l3m3

(xprime3)

|x3|2Y l4m4

(xprime4)

|x4|2Cm1m2m3m4

l1l2l3l4

(minusl1 + l2 + l3 + l4 + 2)


1

)(39)





IS21 =sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

|x2|2Y l3m3

(xprime3)

|x3|2Y l4m4

(xprime4)

|x4|2Cm1m2m3m4

l1l2l3l4

(minusl1 + l2 + l3 + l4 + 2)


1

)IS22 =

sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

|x2|2Y l3m3

(xprime3)

|x3|2Y l4m4

(xprime4)

|x4|2Cm1m2m3m4

l1l2l3l4

(minusl1 + l2 + l3 + l4 + 2)


)(310)




primemicro3 x



I2 = IS21 + IR21 (311)


I3 =sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

int |x3||x2|


=sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4log

(|x3||x2|

)δ(l1 + l2 l3 + l4)

+sum

limil 6=0

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2

(|x3|l minus |x2|l

l


(312)

ndash 11 ndash

JHEP03(2016)165


IL32 =minussumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4log (|x2|) δ(l1 + l2 l3 + l4)

IS32 =sum

limil 6=0

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

(minus|x2|l

l


IL33 =sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4log (|x3|) δ(l1 + l2 l3 + l4)

IS33 =sum

limil 6=0

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

(|x3|l

l


(313)




I3 = IL32 + IS32 + IL33 + IS33 (314)



we know that

|x2| = max(r1 r2) =1

2(r1 + r2 + |r1 minus r2|)

|x3| = min(r3 r4) =1

2(r3 + r4 minus |r3 minus r4|) (315)



)(316)





I4 =sumlimi

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)


int |x4||x3|


=sum

limille0

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)

|x4|2


l minus 2



I43 =sum

limille0

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)

|x4|2

(minus|x3|lminus2

l minus 2


I44 =sum

limille0

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)

|x4|2

(|x4|lminus2

l minus 2

)(318)

ndash 12 ndash

JHEP03(2016)165




I4 = I43 + I44 (319)


I5 =sumlimi

Cm1m2m3m4

l1l2l3l4

prodi

Y limi(xi)

int infin|x4|


=sumlimil

Cm1m2m3m4

l1l2l3l4

prodi

Y limi(xi)

(|x4|minus4minusl

4 + l

)δ(l l1 + l2 + l3 + l4) (320)






C2 = ∆(∆minus 4) + l(l + 2) = minus3 (321)










and Libine [29]


result is

log(r3r2

)sumlimi

(Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

)δ(l1 + l2 l3 + l4) (41)


Φ(s t) = F0(s t) + log(s)F1(s t) (42)



ndash 13 ndash

JHEP03(2016)165





)spart

parts

(spartH

parts

)minus(


(tpartH

partt

)+ 4s

partH

parts(44)


C2 sΦ = 2 sΦ C2 sF1 = 2 sF1 (45)



for the integral is

1

2x213x224

F1(s t) (46)



plies that

2sumlimi

(Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

)δ(l1 + l2 l3 + l4) =

1

x213x224

F1(s t)

(47)




V+ otimes V+ = A[200] +

infinoplusk=1

D[k+2 k2 k2] (48)








P++++ defined by




ndash 14 ndash

JHEP03(2016)165



















1

2x213x224



prime24 P(x1 x2 x










ndash 15 ndash

JHEP03(2016)165





m1(x)Y l2

m2(x)Y l5

m5(x) (416)


(x1)Yl2m2


Y l1m1

(x1)otimes Y l2m2


(x)Y l2m2

(x)

Y l3m3

(xprime3)

|x3|2otimesY l4m4

(xprime4)

|x4|2rarr

Y l3m3

(xprime)

|x|2otimesY l4m4

(xprime)

|x|2(417)















m(K)Y ml (xprime)

sumlimi

(minusi)l12l1 l1

Y l1m1


Y l2m2

(x2)(i)l3

2l3 l3Y l3m3

(xprime3)(i)l4

2l4 l4Y l4m4

(xprime4)

P++minusminus

(Y m1l1




(K)vminus)

=sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

P++minusminus

(Y m1l1

(P )

2l1 l1v+ otimes

Y m2l2

(P )

2l2 l2v+ otimes

Y m3l3

(K)

2l3 l3vminus otimes

Y m4l4

(K)

2l4 l4vminus)

(418)








ndash 16 ndash

JHEP03(2016)165






F1 =1

λlog

((1 + ρs)


)(419)



sk

u2k+1[Qk(u) log(1 + u)]+ (420)



Qk(u) =

ksuml=0

bl(k)ul (421)

where

bl(k) =

(k

k minus l

)(2k + l

k minus l

)(422)








equivariant maps


I = I1 + I2 + I3 + I4 + I5



I(1) = (IS11 + IS21)

I(2) = (IS22 + IS32 + IL32)

I(3) = (IL33 + IS33 + IS43)

I(4) = (IS45 + IS55) (52)

ndash 17 ndash

JHEP03(2016)165




I = I(1) + I(2) + I(3) + I(4) (53)



jI(i) = minus2δij

prodk 6=j

1

(xk minus xi)2(54)









I(1) =sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

r22

Y l3m3

(xprime3)

r23

Y l4m4

(xprime4)



(l1 + l2 + l3 + l4 + 4)(minusl1 + l2 + l3 + l4 + 2)(55)




part2


(r21)A = 4A(A+ 1)(r21)Aminus1

part


part

partxmicro1Y l1m1

(x1)part


21)Aminus1Y l1

m1(x1) (56)


part2


I(1) =sumlimi


r22r23r

24

Y l1m1

(x1)Yl2m2

(xprime2)Yl3m3

(xprime3)Yl4m4


(57)


1


1

r22r23r

24

sumlimi

Y l2m2

(x1)Yl3m3

(x1)Yl4m4

(x1)Yl2m2

(xprime2)Yl3m3

(xprime3)Yl4m4

(xprime4)

=(l1 + 1)rl2+l3+l41

r22r23r

24

sumlimi

Cm1m2m3m4l1l2l3l4

Y l1m1

(x)Y l2m2

(xprime2)Yl3m3

(xprime3)Yl4m4

(xprime4) (58)

ndash 18 ndash

JHEP03(2016)165














I(2) =sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


1

l2+l3+l4minusl1+2

δ(l1 + l2 l3 + l4)

r23r24

+sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


1

r23r24

times



]


part

partxmicro2Y l2m2

(x2)part


22)Aminus1Y l2

m2(x2)

part

partxmicro2logr2 =

xmicro2r22

part

partxmicro2

part

partxmicro2logr2 =

2

r22(59)

We find

part2




r23r24

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


(510)


minus 2







ndash 19 ndash

JHEP03(2016)165


equivariant map


such that



x22x23x

24








23x

24


prime2)f(x1 x

prime3)f(x1 x

prime4)

(515)



1


(516)







part2



prime2)f(x1 x

prime3)f(x1 x

prime4) (518)


24I

(1)(x1 x2 x3 x4)


x22x23x

24 I(1)(x1 x2 x3 x4)rarr F1(e


While x23x24I








ndash 20 ndash

JHEP03(2016)165

implies that (part

partxmicro1+ x22

part

partxmicro2

1

x22+ x23

part

partxmicro3

1

x23+ x24

part

partxmicro4

1

x24

)F1 = 0 (521)


24I



24(I

(1)+I(2)+I(3)+I(4)) = x22x23x



partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)I = 0 (522)





23x

24I(x1 x2 x3 x4) =

x22x23x

24

x213x224

Φ(s t)

(523)



with


minus)=xprime21 x

prime22 xprime23


The x21x22x

23I













〈φ(x1)φ3(x2)φ(x3)φ(x4)〉 (528)




ndash 21 ndash

JHEP03(2016)165



twisted co-product










23x

24



















prime22 xprime23






ndash 22 ndash

JHEP03(2016)165





I2 =

intd4x5d

4x6xminus256

x215x225x

245x

226x

246x

236

(61)






















ndash 23 ndash

JHEP03(2016)165
































ndash 24 ndash

JHEP03(2016)165































Acknowledgments








this work was done

ndash 25 ndash

JHEP03(2016)165








1

2π2

intdS3Y l1

m1(x)Y l2

m2(x) = gm1m2

δl1l2

(l1 + 1)(A1)




gm1m2 = δm1m2


m2(x) = Y l

m1(x)

Y ml (x) = Y l

m(x) (A2)






(mLmR)

gm1m2 = δm1minusm2

Y ml (x) = Y l

minusm(x) (A3)



ables are

z = x1 + ix2 harr(

1

2

1

2


(minus1

2minus1

2

)w = x3 + ix4 harr

(1

2minus1

2


(minus1

21

2

)(A4)

We find that

Y l(l2l2) = (x1 + ix2)

l = zl (A5)

has normalization

1

2π2

intdS3(Y l

( l2 l2))lowast Y l

( l2 l2)

=1

(l + 1)(A6)


ndash 26 ndash

JHEP03(2016)165




(minusi)l

2llY lm(x)Y m

l (P )v+ (A7)


η(Y l1m1

(P )v+ Y m2l2

(K)vminus)

= δl1l2δm1m222l1(l1)2 (A8)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)= δl1l2δm1m2 (A9)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)=

1

2π2

intS3

Y m2l2

(xprime)


m1(x)) (A10)





Y lm(xprime)

xprime2


m(x) (A12)


η = f η


1


summ

1

y2Y lm(x)Y m


=infinsuml=0

summ

1

x2Y lm(xprime)Y m

l (y) if |x| gt |y| (A13)






ndash 27 ndash

JHEP03(2016)165


C2 = minus1

2LMNL


2MpqMqp (A14)






(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)eminusiP middotxv+

=

(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)ipart





Mpq = xppart

partxqminus xq

part

partxp(A16)


1

2MpqMqp(r

AY lm(x)) =

rA

2MpqMqpY


m(x) (A17)





m(x) (A19)

Thus we have

C2 rAY l




m(x) (A21)





F1 =1

λlog

((1 + ρs)


)(B1)

ndash 28 ndash

JHEP03(2016)165


of s to obtain

F1 =log(1+u)

u+

s


s2

u5(3u(2+u)minus 6 + 6u+ u2) log(1+u)

)+s3


)(B2)


F1 =

(minus1 +

u

2minus u2

3+u3


)+ s

(minus1

6+u

6minus 3u2


)+ s2

(minus1

30+

u

20minus 2u2


)+s3

(minus1

140+

u

70minus 5u2



The term at sk is

sk









in Pk(u) is

(2k)

(k)2(B6)


Qk(u) =ksuml=0

blul (B7)


ksuml=0

bl(minus1)l

(iminus l)= 0 (B8)




b1 =5b02 b2 =

20b09 b3 =

5b06 b4 =

5b042 b5 =

b0252

(B9)

ndash 29 ndash

JHEP03(2016)165



ksuml=0

MIlbl = 0 (B10)

where

MIl =(minus1)l




MIlbl (B12)



bl =sumI




bIb0 = 1 16bIb0 = 32 35 120bIb0 = 2 97 27 170

bIb0 =

5

2

20

95

6

5

42

1

252

bIb0 =

3

75

22

20

11

5

11

1

22

1

924

bIb0 =

7

2

63

13175

52

175

143

63

286

7

429

1

3432

(B14)



b1(k)b0 = k2

bk(k)b0 =k2

(2k)

bkminus1(k)b0 =k2

(2k)k(k + 1)

bkminus2(k)b0 =k2

(2k)

(k minus 1)k(k + 1)(k + 2)

4

bkminus3(k)b0 =k2

(2k)

(k

3

)(k + 3

3

)bkminusm(k)b0 =

k2

(2k)

(k

m

)(k +m

m

)(B15)

ndash 30 ndash

JHEP03(2016)165


so that

we have

b1(k) = k2(2k)

k2

bk(k) = 1



4

bkminus3(k) =

(k

3

)(k + 3

3

)bkminusm(k) =

(k

m

)(k +m

m

)(B16)


Pk(u) =

ksumi=1

aiuiminus1 (B17)



bl(minus1)iminusl

(iminus l)= 0 (B18)




k = 2 ~a = a1 ak = 6 3

k = 3 ~a =1

3times 60 60 11

k = 4 ~a =5

6times 84 126 52 5

k = 5 ~a =1

30times 7560 15120 9870 2310 137

k = 6 ~a =7

10times 13720 3300 2960 1140 174 7 (B19)

which leads to

a1(k) =(2k)

(k)2

ak(k) = 2h(k) = 2

ksuml=1

1

l(B20)




ndash 31 ndash

JHEP03(2016)165


grals






infinsumli=0

summi

infinsumt=0

sump


(x1)Yl2minusm2

(x2)Yl3m3

(xprime3)Yl4m4



t+ 1

=x23x

24

x213x224

1

λlog

((1 + ρs)


)(B21)



C l1 l2 lm1m2 m1+m2

=

langl12

l22

mL12

mL22

∣∣∣∣∣ l2

mL1 +mL2

2

ranglangl12

l22

mR12

mR22

∣∣∣∣∣ l2

mR1 +mR22






Yn12

n12n12

(x1)Yn22

n22n22



(x4) = zn11 zn2

2 zn33 zn4

4 (B22)


zn11 zn2

2 zn33 zn4


note that





that zn11 zn2

2 zn33 zn4


Yn12

n12n1minus2

2

(x1)Yn22

n2minus22

n22

(x2)Yn32


2

(x3)Yn42

minusn4minus22

minusn42

(x4)

=radicn1n2n3n4z

n1minus11 w1z

n2minus12 w2z

n3minus13 w3z

n4minus14 w4 (B24)


coefficients

Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B25)

Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B26)

ndash 32 ndash

JHEP03(2016)165


Cn12n22n1+n2

2n12n2minus2

2n1+n2minus2

2

=

radicn2

n1 + n2(B27)

Cn12n22n1+n2minus2

2n12n2minus2

2n1+n2minus2

2

= minusradic

n1n1 + n2

(B28)



n2minus12 w2z

n3minus13 w3z

n4minus14 w4 is

radicn1n2n3n4

[Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2 + 1

+Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2

]=


(B29)




section 5






parth








prime4vminus) (C1)



(2ixprime3microx

prime3 middot

part


part


)eiKmiddotx

prime3vminus

= ix23part

partxmicro3

1

x23eiKmiddotx

prime3vminus (C2)


parth


part

partxmicro2

(1

x22h

)minus x23

part

partxmicro3

(1

x23h

)minus x24

part

partxmicro4

(1

x24h

)(C3)

ndash 33 ndash

JHEP03(2016)165


x22x23x

24

(part

partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)[h

x22x23x

24

]= 0 (C4)


x1micropart

partx1νminus x1ν

part

partx1micro

)h= F1(Mmicroνe









x22x23x

24

(x1micro

part

partx1νminus x1ν

part


part

partx2νminus x2ν

part


part

partx3νminus x3ν

part

partx3micro

+x4micropart

partx4νminus x4ν

part

partx4micro

)[h

x22x23x

24

]= 0 (C6)


consider(x1 middot

part

partx1+ 3

)h = F1(e






x22x23x

24

(x1 middot

part

partx1+ x2 middot

part

partx2+ x3 middot

part

partx3+ x4 middot

part

partx4+ 12

)[h

x22x23x

24

]= 0 (C8)


h

x22x23x

24

=A




we have




ndash 34 ndash

JHEP03(2016)165





infinsumnm=0

im(minus1)n




where







minus (C14)

we find


and

infinsumn=0

(minus1)n




1


(C16)



=infinsum

n1n3n4=0

1



slot we get

α(P middot x2) =1


ndash 35 ndash

JHEP03(2016)165




=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)


=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)

(n1 + n3 + n4)



=infinsumn1

(minus1)n1




timesinfinsumn3

(minus1)n3




timesinfinsumn4

(minus1)n4





=1


1


1


=x22x

33x

24






References














ndash 36 ndash

JHEP03(2016)165










(2003) 013 [hep-th0212208] [INSPIRE]




(2014) 302 [arXiv14036646] [INSPIRE]






















(2015) 006 [arXiv150208005] [INSPIRE]






ndash 37 ndash

JHEP03(2016)165



[INSPIRE]

















(2015) 29FT01 [arXiv150500963] [INSPIRE]








[INSPIRE]














ndash 38 ndash

JHEP03(2016)165















ndash 39 ndash

Introduction























JHEP03(2016)165


























Y lm(P )v+ (13)








Y lm(K)vminus (14)



ndash 2 ndash

JHEP03(2016)165



Φ(x) =1radic2




x2 It is found that

η(Φ(x1)Φ(x2)) =1

(x1 minus x2)2(16)


































ndash 3 ndash

JHEP03(2016)165











































ndash 4 ndash

JHEP03(2016)165















integrals














I(x1 x2 x3 x4) =

intd4x

2π21
















ndash 5 ndash

JHEP03(2016)165







prime4vminus (24)

To this we tensor

































ndash 6 ndash

JHEP03(2016)165


I(x1 x2 x3 x4) =1

2x213x224

Φ(s t) (26)

where

Φ(s t) =1

λ


(t

s

)ln

(1 + ρt

1 + ρs

)+π2

3

)(27)

and

ρ =2



s =x212x

234

x213x224

t =x214x

223

x213x224

(28)



λ sim u+ s

(minus1minus 2

u

)+ s2

(minus 2

u3minus 2

u2

)sim u+ s

(minus1minus 2

u

)ρ sim s2

(2

u5+

1

u4

)+

s


1


s(29)



6minus 1

2log2(minusz) (210)


Φ(s t) =1

λ


(st

)log


(1 + ρs)

)(211)


notation


F0(s u) =1

λ



(1 + ρs)

)F1(s u) = minus 1

λlog


(1 + ρs)









=f(x1 x

prime4)f(x2 x

prime3)


ndash 7 ndash

JHEP03(2016)165

where








in section 4









J3

∣∣∣minus 1

2

rang= minus1

2

∣∣∣minus 1

2

rangJminus

∣∣∣minus 1

2

rang= 0 (214)





JminusJ2+

∣∣∣minus1

2

rang= 0 (215)




(2)

minus 12

The quotient

space Vminus 12V

(2)

minus 12


sequence

0rarr V(2)

minus 12


2V

(2)

minus 12

rarr 0 (216)









ndash 8 ndash

JHEP03(2016)165










(p2)+ rarr 0 (220)
























in section 5


I1 =sumllimi

prod4i=1 Y

limi(x

primei)

|x1|2|x2|2|x3|2|x4|2

int |x1|0


4prodi=1

Y mili

(x) (31)

ndash 9 ndash

JHEP03(2016)165

where d3x = 12π2dS



write it as

Cm1m2m3m4

l1l2l3l4=

suml5m5m6

Cm1m2l5l1l2m5

1

(l5 + 1)Cm3m4l5l3l4m6


(32)

where

Y m1l1

Y m2l2

=suml5m5

Cm1m2l5l1l2m5

Y m5l5

(33)




or equivalently












I1 =sumlimi

prodi Y

limi(x

primei)

|x1|2|x2|2|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

|x1|4+l1+l2+l3+l44 + l1 + l2 + l3 + l4

(38)




ndash 10 ndash

JHEP03(2016)165


region is

I2 =sumlimi

Y l1m1

(x1)prodi=2 Y

limi(x

primei)

|x2|2|x3|2|x4|2

int |x2||x1|


l1(xprime)Y m2

l2(x)Y m3

l3(x)Y m4

l4(x)

=sumlimi

Y l1m1

(x1)prodi=2 Y

limi(x

primei)

|x2|2|x3|2|x4|2

int |x2||x1|


l1l2l3l4

=sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

|x2|2Y l3m3

(xprime3)

|x3|2Y l4m4

(xprime4)

|x4|2Cm1m2m3m4

l1l2l3l4

(minusl1 + l2 + l3 + l4 + 2)


1

)(39)





IS21 =sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

|x2|2Y l3m3

(xprime3)

|x3|2Y l4m4

(xprime4)

|x4|2Cm1m2m3m4

l1l2l3l4

(minusl1 + l2 + l3 + l4 + 2)


1

)IS22 =

sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

|x2|2Y l3m3

(xprime3)

|x3|2Y l4m4

(xprime4)

|x4|2Cm1m2m3m4

l1l2l3l4

(minusl1 + l2 + l3 + l4 + 2)


)(310)




primemicro3 x



I2 = IS21 + IR21 (311)


I3 =sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

int |x3||x2|


=sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4log

(|x3||x2|

)δ(l1 + l2 l3 + l4)

+sum

limil 6=0

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2

(|x3|l minus |x2|l

l


(312)

ndash 11 ndash

JHEP03(2016)165


IL32 =minussumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4log (|x2|) δ(l1 + l2 l3 + l4)

IS32 =sum

limil 6=0

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

(minus|x2|l

l


IL33 =sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4log (|x3|) δ(l1 + l2 l3 + l4)

IS33 =sum

limil 6=0

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

(|x3|l

l


(313)




I3 = IL32 + IS32 + IL33 + IS33 (314)



we know that

|x2| = max(r1 r2) =1

2(r1 + r2 + |r1 minus r2|)

|x3| = min(r3 r4) =1

2(r3 + r4 minus |r3 minus r4|) (315)



)(316)





I4 =sumlimi

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)


int |x4||x3|


=sum

limille0

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)

|x4|2


l minus 2



I43 =sum

limille0

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)

|x4|2

(minus|x3|lminus2

l minus 2


I44 =sum

limille0

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)

|x4|2

(|x4|lminus2

l minus 2

)(318)

ndash 12 ndash

JHEP03(2016)165




I4 = I43 + I44 (319)


I5 =sumlimi

Cm1m2m3m4

l1l2l3l4

prodi

Y limi(xi)

int infin|x4|


=sumlimil

Cm1m2m3m4

l1l2l3l4

prodi

Y limi(xi)

(|x4|minus4minusl

4 + l

)δ(l l1 + l2 + l3 + l4) (320)






C2 = ∆(∆minus 4) + l(l + 2) = minus3 (321)










and Libine [29]


result is

log(r3r2

)sumlimi

(Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

)δ(l1 + l2 l3 + l4) (41)


Φ(s t) = F0(s t) + log(s)F1(s t) (42)



ndash 13 ndash

JHEP03(2016)165





)spart

parts

(spartH

parts

)minus(


(tpartH

partt

)+ 4s

partH

parts(44)


C2 sΦ = 2 sΦ C2 sF1 = 2 sF1 (45)



for the integral is

1

2x213x224

F1(s t) (46)



plies that

2sumlimi

(Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

)δ(l1 + l2 l3 + l4) =

1

x213x224

F1(s t)

(47)




V+ otimes V+ = A[200] +

infinoplusk=1

D[k+2 k2 k2] (48)








P++++ defined by




ndash 14 ndash

JHEP03(2016)165



















1

2x213x224



prime24 P(x1 x2 x










ndash 15 ndash

JHEP03(2016)165





m1(x)Y l2

m2(x)Y l5

m5(x) (416)


(x1)Yl2m2


Y l1m1

(x1)otimes Y l2m2


(x)Y l2m2

(x)

Y l3m3

(xprime3)

|x3|2otimesY l4m4

(xprime4)

|x4|2rarr

Y l3m3

(xprime)

|x|2otimesY l4m4

(xprime)

|x|2(417)















m(K)Y ml (xprime)

sumlimi

(minusi)l12l1 l1

Y l1m1


Y l2m2

(x2)(i)l3

2l3 l3Y l3m3

(xprime3)(i)l4

2l4 l4Y l4m4

(xprime4)

P++minusminus

(Y m1l1




(K)vminus)

=sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

P++minusminus

(Y m1l1

(P )

2l1 l1v+ otimes

Y m2l2

(P )

2l2 l2v+ otimes

Y m3l3

(K)

2l3 l3vminus otimes

Y m4l4

(K)

2l4 l4vminus)

(418)








ndash 16 ndash

JHEP03(2016)165






F1 =1

λlog

((1 + ρs)


)(419)



sk

u2k+1[Qk(u) log(1 + u)]+ (420)



Qk(u) =

ksuml=0

bl(k)ul (421)

where

bl(k) =

(k

k minus l

)(2k + l

k minus l

)(422)








equivariant maps


I = I1 + I2 + I3 + I4 + I5



I(1) = (IS11 + IS21)

I(2) = (IS22 + IS32 + IL32)

I(3) = (IL33 + IS33 + IS43)

I(4) = (IS45 + IS55) (52)

ndash 17 ndash

JHEP03(2016)165




I = I(1) + I(2) + I(3) + I(4) (53)



jI(i) = minus2δij

prodk 6=j

1

(xk minus xi)2(54)









I(1) =sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

r22

Y l3m3

(xprime3)

r23

Y l4m4

(xprime4)



(l1 + l2 + l3 + l4 + 4)(minusl1 + l2 + l3 + l4 + 2)(55)




part2


(r21)A = 4A(A+ 1)(r21)Aminus1

part


part

partxmicro1Y l1m1

(x1)part


21)Aminus1Y l1

m1(x1) (56)


part2


I(1) =sumlimi


r22r23r

24

Y l1m1

(x1)Yl2m2

(xprime2)Yl3m3

(xprime3)Yl4m4


(57)


1


1

r22r23r

24

sumlimi

Y l2m2

(x1)Yl3m3

(x1)Yl4m4

(x1)Yl2m2

(xprime2)Yl3m3

(xprime3)Yl4m4

(xprime4)

=(l1 + 1)rl2+l3+l41

r22r23r

24

sumlimi

Cm1m2m3m4l1l2l3l4

Y l1m1

(x)Y l2m2

(xprime2)Yl3m3

(xprime3)Yl4m4

(xprime4) (58)

ndash 18 ndash

JHEP03(2016)165














I(2) =sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


1

l2+l3+l4minusl1+2

δ(l1 + l2 l3 + l4)

r23r24

+sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


1

r23r24

times



]


part

partxmicro2Y l2m2

(x2)part


22)Aminus1Y l2

m2(x2)

part

partxmicro2logr2 =

xmicro2r22

part

partxmicro2

part

partxmicro2logr2 =

2

r22(59)

We find

part2




r23r24

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


(510)


minus 2







ndash 19 ndash

JHEP03(2016)165


equivariant map


such that



x22x23x

24








23x

24


prime2)f(x1 x

prime3)f(x1 x

prime4)

(515)



1


(516)







part2



prime2)f(x1 x

prime3)f(x1 x

prime4) (518)


24I

(1)(x1 x2 x3 x4)


x22x23x

24 I(1)(x1 x2 x3 x4)rarr F1(e


While x23x24I








ndash 20 ndash

JHEP03(2016)165

implies that (part

partxmicro1+ x22

part

partxmicro2

1

x22+ x23

part

partxmicro3

1

x23+ x24

part

partxmicro4

1

x24

)F1 = 0 (521)


24I



24(I

(1)+I(2)+I(3)+I(4)) = x22x23x



partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)I = 0 (522)





23x

24I(x1 x2 x3 x4) =

x22x23x

24

x213x224

Φ(s t)

(523)



with


minus)=xprime21 x

prime22 xprime23


The x21x22x

23I













〈φ(x1)φ3(x2)φ(x3)φ(x4)〉 (528)




ndash 21 ndash

JHEP03(2016)165



twisted co-product










23x

24



















prime22 xprime23






ndash 22 ndash

JHEP03(2016)165





I2 =

intd4x5d

4x6xminus256

x215x225x

245x

226x

246x

236

(61)






















ndash 23 ndash

JHEP03(2016)165
































ndash 24 ndash

JHEP03(2016)165































Acknowledgments








this work was done

ndash 25 ndash

JHEP03(2016)165








1

2π2

intdS3Y l1

m1(x)Y l2

m2(x) = gm1m2

δl1l2

(l1 + 1)(A1)




gm1m2 = δm1m2


m2(x) = Y l

m1(x)

Y ml (x) = Y l

m(x) (A2)






(mLmR)

gm1m2 = δm1minusm2

Y ml (x) = Y l

minusm(x) (A3)



ables are

z = x1 + ix2 harr(

1

2

1

2


(minus1

2minus1

2

)w = x3 + ix4 harr

(1

2minus1

2


(minus1

21

2

)(A4)

We find that

Y l(l2l2) = (x1 + ix2)

l = zl (A5)

has normalization

1

2π2

intdS3(Y l

( l2 l2))lowast Y l

( l2 l2)

=1

(l + 1)(A6)


ndash 26 ndash

JHEP03(2016)165




(minusi)l

2llY lm(x)Y m

l (P )v+ (A7)


η(Y l1m1

(P )v+ Y m2l2

(K)vminus)

= δl1l2δm1m222l1(l1)2 (A8)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)= δl1l2δm1m2 (A9)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)=

1

2π2

intS3

Y m2l2

(xprime)


m1(x)) (A10)





Y lm(xprime)

xprime2


m(x) (A12)


η = f η


1


summ

1

y2Y lm(x)Y m


=infinsuml=0

summ

1

x2Y lm(xprime)Y m

l (y) if |x| gt |y| (A13)






ndash 27 ndash

JHEP03(2016)165


C2 = minus1

2LMNL


2MpqMqp (A14)






(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)eminusiP middotxv+

=

(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)ipart





Mpq = xppart

partxqminus xq

part

partxp(A16)


1

2MpqMqp(r

AY lm(x)) =

rA

2MpqMqpY


m(x) (A17)





m(x) (A19)

Thus we have

C2 rAY l




m(x) (A21)





F1 =1

λlog

((1 + ρs)


)(B1)

ndash 28 ndash

JHEP03(2016)165


of s to obtain

F1 =log(1+u)

u+

s


s2

u5(3u(2+u)minus 6 + 6u+ u2) log(1+u)

)+s3


)(B2)


F1 =

(minus1 +

u

2minus u2

3+u3


)+ s

(minus1

6+u

6minus 3u2


)+ s2

(minus1

30+

u

20minus 2u2


)+s3

(minus1

140+

u

70minus 5u2



The term at sk is

sk









in Pk(u) is

(2k)

(k)2(B6)


Qk(u) =ksuml=0

blul (B7)


ksuml=0

bl(minus1)l

(iminus l)= 0 (B8)




b1 =5b02 b2 =

20b09 b3 =

5b06 b4 =

5b042 b5 =

b0252

(B9)

ndash 29 ndash

JHEP03(2016)165



ksuml=0

MIlbl = 0 (B10)

where

MIl =(minus1)l




MIlbl (B12)



bl =sumI




bIb0 = 1 16bIb0 = 32 35 120bIb0 = 2 97 27 170

bIb0 =

5

2

20

95

6

5

42

1

252

bIb0 =

3

75

22

20

11

5

11

1

22

1

924

bIb0 =

7

2

63

13175

52

175

143

63

286

7

429

1

3432

(B14)



b1(k)b0 = k2

bk(k)b0 =k2

(2k)

bkminus1(k)b0 =k2

(2k)k(k + 1)

bkminus2(k)b0 =k2

(2k)

(k minus 1)k(k + 1)(k + 2)

4

bkminus3(k)b0 =k2

(2k)

(k

3

)(k + 3

3

)bkminusm(k)b0 =

k2

(2k)

(k

m

)(k +m

m

)(B15)

ndash 30 ndash

JHEP03(2016)165


so that

we have

b1(k) = k2(2k)

k2

bk(k) = 1



4

bkminus3(k) =

(k

3

)(k + 3

3

)bkminusm(k) =

(k

m

)(k +m

m

)(B16)


Pk(u) =

ksumi=1

aiuiminus1 (B17)



bl(minus1)iminusl

(iminus l)= 0 (B18)




k = 2 ~a = a1 ak = 6 3

k = 3 ~a =1

3times 60 60 11

k = 4 ~a =5

6times 84 126 52 5

k = 5 ~a =1

30times 7560 15120 9870 2310 137

k = 6 ~a =7

10times 13720 3300 2960 1140 174 7 (B19)

which leads to

a1(k) =(2k)

(k)2

ak(k) = 2h(k) = 2

ksuml=1

1

l(B20)




ndash 31 ndash

JHEP03(2016)165


grals






infinsumli=0

summi

infinsumt=0

sump


(x1)Yl2minusm2

(x2)Yl3m3

(xprime3)Yl4m4



t+ 1

=x23x

24

x213x224

1

λlog

((1 + ρs)


)(B21)



C l1 l2 lm1m2 m1+m2

=

langl12

l22

mL12

mL22

∣∣∣∣∣ l2

mL1 +mL2

2

ranglangl12

l22

mR12

mR22

∣∣∣∣∣ l2

mR1 +mR22






Yn12

n12n12

(x1)Yn22

n22n22



(x4) = zn11 zn2

2 zn33 zn4

4 (B22)


zn11 zn2

2 zn33 zn4


note that





that zn11 zn2

2 zn33 zn4


Yn12

n12n1minus2

2

(x1)Yn22

n2minus22

n22

(x2)Yn32


2

(x3)Yn42

minusn4minus22

minusn42

(x4)

=radicn1n2n3n4z

n1minus11 w1z

n2minus12 w2z

n3minus13 w3z

n4minus14 w4 (B24)


coefficients

Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B25)

Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B26)

ndash 32 ndash

JHEP03(2016)165


Cn12n22n1+n2

2n12n2minus2

2n1+n2minus2

2

=

radicn2

n1 + n2(B27)

Cn12n22n1+n2minus2

2n12n2minus2

2n1+n2minus2

2

= minusradic

n1n1 + n2

(B28)



n2minus12 w2z

n3minus13 w3z

n4minus14 w4 is

radicn1n2n3n4

[Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2 + 1

+Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2

]=


(B29)




section 5






parth








prime4vminus) (C1)



(2ixprime3microx

prime3 middot

part


part


)eiKmiddotx

prime3vminus

= ix23part

partxmicro3

1

x23eiKmiddotx

prime3vminus (C2)


parth


part

partxmicro2

(1

x22h

)minus x23

part

partxmicro3

(1

x23h

)minus x24

part

partxmicro4

(1

x24h

)(C3)

ndash 33 ndash

JHEP03(2016)165


x22x23x

24

(part

partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)[h

x22x23x

24

]= 0 (C4)


x1micropart

partx1νminus x1ν

part

partx1micro

)h= F1(Mmicroνe









x22x23x

24

(x1micro

part

partx1νminus x1ν

part


part

partx2νminus x2ν

part


part

partx3νminus x3ν

part

partx3micro

+x4micropart

partx4νminus x4ν

part

partx4micro

)[h

x22x23x

24

]= 0 (C6)


consider(x1 middot

part

partx1+ 3

)h = F1(e






x22x23x

24

(x1 middot

part

partx1+ x2 middot

part

partx2+ x3 middot

part

partx3+ x4 middot

part

partx4+ 12

)[h

x22x23x

24

]= 0 (C8)


h

x22x23x

24

=A




we have




ndash 34 ndash

JHEP03(2016)165





infinsumnm=0

im(minus1)n




where







minus (C14)

we find


and

infinsumn=0

(minus1)n




1


(C16)



=infinsum

n1n3n4=0

1



slot we get

α(P middot x2) =1


ndash 35 ndash

JHEP03(2016)165




=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)


=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)

(n1 + n3 + n4)



=infinsumn1

(minus1)n1




timesinfinsumn3

(minus1)n3




timesinfinsumn4

(minus1)n4





=1


1


1


=x22x

33x

24






References














ndash 36 ndash

JHEP03(2016)165










(2003) 013 [hep-th0212208] [INSPIRE]




(2014) 302 [arXiv14036646] [INSPIRE]






















(2015) 006 [arXiv150208005] [INSPIRE]






ndash 37 ndash

JHEP03(2016)165



[INSPIRE]

















(2015) 29FT01 [arXiv150500963] [INSPIRE]








[INSPIRE]














ndash 38 ndash

JHEP03(2016)165















ndash 39 ndash

Introduction























JHEP03(2016)165



Φ(x) =1radic2




x2 It is found that

η(Φ(x1)Φ(x2)) =1

(x1 minus x2)2(16)


































ndash 3 ndash

JHEP03(2016)165











































ndash 4 ndash

JHEP03(2016)165















integrals














I(x1 x2 x3 x4) =

intd4x

2π21
















ndash 5 ndash

JHEP03(2016)165







prime4vminus (24)

To this we tensor

































ndash 6 ndash

JHEP03(2016)165


I(x1 x2 x3 x4) =1

2x213x224

Φ(s t) (26)

where

Φ(s t) =1

λ


(t

s

)ln

(1 + ρt

1 + ρs

)+π2

3

)(27)

and

ρ =2



s =x212x

234

x213x224

t =x214x

223

x213x224

(28)



λ sim u+ s

(minus1minus 2

u

)+ s2

(minus 2

u3minus 2

u2

)sim u+ s

(minus1minus 2

u

)ρ sim s2

(2

u5+

1

u4

)+

s


1


s(29)



6minus 1

2log2(minusz) (210)


Φ(s t) =1

λ


(st

)log


(1 + ρs)

)(211)


notation


F0(s u) =1

λ



(1 + ρs)

)F1(s u) = minus 1

λlog


(1 + ρs)









=f(x1 x

prime4)f(x2 x

prime3)


ndash 7 ndash

JHEP03(2016)165

where








in section 4









J3

∣∣∣minus 1

2

rang= minus1

2

∣∣∣minus 1

2

rangJminus

∣∣∣minus 1

2

rang= 0 (214)





JminusJ2+

∣∣∣minus1

2

rang= 0 (215)




(2)

minus 12

The quotient

space Vminus 12V

(2)

minus 12


sequence

0rarr V(2)

minus 12


2V

(2)

minus 12

rarr 0 (216)









ndash 8 ndash

JHEP03(2016)165










(p2)+ rarr 0 (220)
























in section 5


I1 =sumllimi

prod4i=1 Y

limi(x

primei)

|x1|2|x2|2|x3|2|x4|2

int |x1|0


4prodi=1

Y mili

(x) (31)

ndash 9 ndash

JHEP03(2016)165

where d3x = 12π2dS



write it as

Cm1m2m3m4

l1l2l3l4=

suml5m5m6

Cm1m2l5l1l2m5

1

(l5 + 1)Cm3m4l5l3l4m6


(32)

where

Y m1l1

Y m2l2

=suml5m5

Cm1m2l5l1l2m5

Y m5l5

(33)




or equivalently












I1 =sumlimi

prodi Y

limi(x

primei)

|x1|2|x2|2|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

|x1|4+l1+l2+l3+l44 + l1 + l2 + l3 + l4

(38)




ndash 10 ndash

JHEP03(2016)165


region is

I2 =sumlimi

Y l1m1

(x1)prodi=2 Y

limi(x

primei)

|x2|2|x3|2|x4|2

int |x2||x1|


l1(xprime)Y m2

l2(x)Y m3

l3(x)Y m4

l4(x)

=sumlimi

Y l1m1

(x1)prodi=2 Y

limi(x

primei)

|x2|2|x3|2|x4|2

int |x2||x1|


l1l2l3l4

=sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

|x2|2Y l3m3

(xprime3)

|x3|2Y l4m4

(xprime4)

|x4|2Cm1m2m3m4

l1l2l3l4

(minusl1 + l2 + l3 + l4 + 2)


1

)(39)





IS21 =sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

|x2|2Y l3m3

(xprime3)

|x3|2Y l4m4

(xprime4)

|x4|2Cm1m2m3m4

l1l2l3l4

(minusl1 + l2 + l3 + l4 + 2)


1

)IS22 =

sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

|x2|2Y l3m3

(xprime3)

|x3|2Y l4m4

(xprime4)

|x4|2Cm1m2m3m4

l1l2l3l4

(minusl1 + l2 + l3 + l4 + 2)


)(310)




primemicro3 x



I2 = IS21 + IR21 (311)


I3 =sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

int |x3||x2|


=sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4log

(|x3||x2|

)δ(l1 + l2 l3 + l4)

+sum

limil 6=0

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2

(|x3|l minus |x2|l

l


(312)

ndash 11 ndash

JHEP03(2016)165


IL32 =minussumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4log (|x2|) δ(l1 + l2 l3 + l4)

IS32 =sum

limil 6=0

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

(minus|x2|l

l


IL33 =sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4log (|x3|) δ(l1 + l2 l3 + l4)

IS33 =sum

limil 6=0

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

(|x3|l

l


(313)




I3 = IL32 + IS32 + IL33 + IS33 (314)



we know that

|x2| = max(r1 r2) =1

2(r1 + r2 + |r1 minus r2|)

|x3| = min(r3 r4) =1

2(r3 + r4 minus |r3 minus r4|) (315)



)(316)





I4 =sumlimi

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)


int |x4||x3|


=sum

limille0

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)

|x4|2


l minus 2



I43 =sum

limille0

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)

|x4|2

(minus|x3|lminus2

l minus 2


I44 =sum

limille0

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)

|x4|2

(|x4|lminus2

l minus 2

)(318)

ndash 12 ndash

JHEP03(2016)165




I4 = I43 + I44 (319)


I5 =sumlimi

Cm1m2m3m4

l1l2l3l4

prodi

Y limi(xi)

int infin|x4|


=sumlimil

Cm1m2m3m4

l1l2l3l4

prodi

Y limi(xi)

(|x4|minus4minusl

4 + l

)δ(l l1 + l2 + l3 + l4) (320)






C2 = ∆(∆minus 4) + l(l + 2) = minus3 (321)










and Libine [29]


result is

log(r3r2

)sumlimi

(Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

)δ(l1 + l2 l3 + l4) (41)


Φ(s t) = F0(s t) + log(s)F1(s t) (42)



ndash 13 ndash

JHEP03(2016)165





)spart

parts

(spartH

parts

)minus(


(tpartH

partt

)+ 4s

partH

parts(44)


C2 sΦ = 2 sΦ C2 sF1 = 2 sF1 (45)



for the integral is

1

2x213x224

F1(s t) (46)



plies that

2sumlimi

(Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

)δ(l1 + l2 l3 + l4) =

1

x213x224

F1(s t)

(47)




V+ otimes V+ = A[200] +

infinoplusk=1

D[k+2 k2 k2] (48)








P++++ defined by




ndash 14 ndash

JHEP03(2016)165



















1

2x213x224



prime24 P(x1 x2 x










ndash 15 ndash

JHEP03(2016)165





m1(x)Y l2

m2(x)Y l5

m5(x) (416)


(x1)Yl2m2


Y l1m1

(x1)otimes Y l2m2


(x)Y l2m2

(x)

Y l3m3

(xprime3)

|x3|2otimesY l4m4

(xprime4)

|x4|2rarr

Y l3m3

(xprime)

|x|2otimesY l4m4

(xprime)

|x|2(417)















m(K)Y ml (xprime)

sumlimi

(minusi)l12l1 l1

Y l1m1


Y l2m2

(x2)(i)l3

2l3 l3Y l3m3

(xprime3)(i)l4

2l4 l4Y l4m4

(xprime4)

P++minusminus

(Y m1l1




(K)vminus)

=sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

P++minusminus

(Y m1l1

(P )

2l1 l1v+ otimes

Y m2l2

(P )

2l2 l2v+ otimes

Y m3l3

(K)

2l3 l3vminus otimes

Y m4l4

(K)

2l4 l4vminus)

(418)








ndash 16 ndash

JHEP03(2016)165






F1 =1

λlog

((1 + ρs)


)(419)



sk

u2k+1[Qk(u) log(1 + u)]+ (420)



Qk(u) =

ksuml=0

bl(k)ul (421)

where

bl(k) =

(k

k minus l

)(2k + l

k minus l

)(422)








equivariant maps


I = I1 + I2 + I3 + I4 + I5



I(1) = (IS11 + IS21)

I(2) = (IS22 + IS32 + IL32)

I(3) = (IL33 + IS33 + IS43)

I(4) = (IS45 + IS55) (52)

ndash 17 ndash

JHEP03(2016)165




I = I(1) + I(2) + I(3) + I(4) (53)



jI(i) = minus2δij

prodk 6=j

1

(xk minus xi)2(54)









I(1) =sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

r22

Y l3m3

(xprime3)

r23

Y l4m4

(xprime4)



(l1 + l2 + l3 + l4 + 4)(minusl1 + l2 + l3 + l4 + 2)(55)




part2


(r21)A = 4A(A+ 1)(r21)Aminus1

part


part

partxmicro1Y l1m1

(x1)part


21)Aminus1Y l1

m1(x1) (56)


part2


I(1) =sumlimi


r22r23r

24

Y l1m1

(x1)Yl2m2

(xprime2)Yl3m3

(xprime3)Yl4m4


(57)


1


1

r22r23r

24

sumlimi

Y l2m2

(x1)Yl3m3

(x1)Yl4m4

(x1)Yl2m2

(xprime2)Yl3m3

(xprime3)Yl4m4

(xprime4)

=(l1 + 1)rl2+l3+l41

r22r23r

24

sumlimi

Cm1m2m3m4l1l2l3l4

Y l1m1

(x)Y l2m2

(xprime2)Yl3m3

(xprime3)Yl4m4

(xprime4) (58)

ndash 18 ndash

JHEP03(2016)165














I(2) =sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


1

l2+l3+l4minusl1+2

δ(l1 + l2 l3 + l4)

r23r24

+sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


1

r23r24

times



]


part

partxmicro2Y l2m2

(x2)part


22)Aminus1Y l2

m2(x2)

part

partxmicro2logr2 =

xmicro2r22

part

partxmicro2

part

partxmicro2logr2 =

2

r22(59)

We find

part2




r23r24

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


(510)


minus 2







ndash 19 ndash

JHEP03(2016)165


equivariant map


such that



x22x23x

24








23x

24


prime2)f(x1 x

prime3)f(x1 x

prime4)

(515)



1


(516)







part2



prime2)f(x1 x

prime3)f(x1 x

prime4) (518)


24I

(1)(x1 x2 x3 x4)


x22x23x

24 I(1)(x1 x2 x3 x4)rarr F1(e


While x23x24I








ndash 20 ndash

JHEP03(2016)165

implies that (part

partxmicro1+ x22

part

partxmicro2

1

x22+ x23

part

partxmicro3

1

x23+ x24

part

partxmicro4

1

x24

)F1 = 0 (521)


24I



24(I

(1)+I(2)+I(3)+I(4)) = x22x23x



partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)I = 0 (522)





23x

24I(x1 x2 x3 x4) =

x22x23x

24

x213x224

Φ(s t)

(523)



with


minus)=xprime21 x

prime22 xprime23


The x21x22x

23I













〈φ(x1)φ3(x2)φ(x3)φ(x4)〉 (528)




ndash 21 ndash

JHEP03(2016)165



twisted co-product










23x

24



















prime22 xprime23






ndash 22 ndash

JHEP03(2016)165





I2 =

intd4x5d

4x6xminus256

x215x225x

245x

226x

246x

236

(61)






















ndash 23 ndash

JHEP03(2016)165
































ndash 24 ndash

JHEP03(2016)165































Acknowledgments








this work was done

ndash 25 ndash

JHEP03(2016)165








1

2π2

intdS3Y l1

m1(x)Y l2

m2(x) = gm1m2

δl1l2

(l1 + 1)(A1)




gm1m2 = δm1m2


m2(x) = Y l

m1(x)

Y ml (x) = Y l

m(x) (A2)






(mLmR)

gm1m2 = δm1minusm2

Y ml (x) = Y l

minusm(x) (A3)



ables are

z = x1 + ix2 harr(

1

2

1

2


(minus1

2minus1

2

)w = x3 + ix4 harr

(1

2minus1

2


(minus1

21

2

)(A4)

We find that

Y l(l2l2) = (x1 + ix2)

l = zl (A5)

has normalization

1

2π2

intdS3(Y l

( l2 l2))lowast Y l

( l2 l2)

=1

(l + 1)(A6)


ndash 26 ndash

JHEP03(2016)165




(minusi)l

2llY lm(x)Y m

l (P )v+ (A7)


η(Y l1m1

(P )v+ Y m2l2

(K)vminus)

= δl1l2δm1m222l1(l1)2 (A8)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)= δl1l2δm1m2 (A9)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)=

1

2π2

intS3

Y m2l2

(xprime)


m1(x)) (A10)





Y lm(xprime)

xprime2


m(x) (A12)


η = f η


1


summ

1

y2Y lm(x)Y m


=infinsuml=0

summ

1

x2Y lm(xprime)Y m

l (y) if |x| gt |y| (A13)






ndash 27 ndash

JHEP03(2016)165


C2 = minus1

2LMNL


2MpqMqp (A14)






(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)eminusiP middotxv+

=

(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)ipart





Mpq = xppart

partxqminus xq

part

partxp(A16)


1

2MpqMqp(r

AY lm(x)) =

rA

2MpqMqpY


m(x) (A17)





m(x) (A19)

Thus we have

C2 rAY l




m(x) (A21)





F1 =1

λlog

((1 + ρs)


)(B1)

ndash 28 ndash

JHEP03(2016)165


of s to obtain

F1 =log(1+u)

u+

s


s2

u5(3u(2+u)minus 6 + 6u+ u2) log(1+u)

)+s3


)(B2)


F1 =

(minus1 +

u

2minus u2

3+u3


)+ s

(minus1

6+u

6minus 3u2


)+ s2

(minus1

30+

u

20minus 2u2


)+s3

(minus1

140+

u

70minus 5u2



The term at sk is

sk









in Pk(u) is

(2k)

(k)2(B6)


Qk(u) =ksuml=0

blul (B7)


ksuml=0

bl(minus1)l

(iminus l)= 0 (B8)




b1 =5b02 b2 =

20b09 b3 =

5b06 b4 =

5b042 b5 =

b0252

(B9)

ndash 29 ndash

JHEP03(2016)165



ksuml=0

MIlbl = 0 (B10)

where

MIl =(minus1)l




MIlbl (B12)



bl =sumI




bIb0 = 1 16bIb0 = 32 35 120bIb0 = 2 97 27 170

bIb0 =

5

2

20

95

6

5

42

1

252

bIb0 =

3

75

22

20

11

5

11

1

22

1

924

bIb0 =

7

2

63

13175

52

175

143

63

286

7

429

1

3432

(B14)



b1(k)b0 = k2

bk(k)b0 =k2

(2k)

bkminus1(k)b0 =k2

(2k)k(k + 1)

bkminus2(k)b0 =k2

(2k)

(k minus 1)k(k + 1)(k + 2)

4

bkminus3(k)b0 =k2

(2k)

(k

3

)(k + 3

3

)bkminusm(k)b0 =

k2

(2k)

(k

m

)(k +m

m

)(B15)

ndash 30 ndash

JHEP03(2016)165


so that

we have

b1(k) = k2(2k)

k2

bk(k) = 1



4

bkminus3(k) =

(k

3

)(k + 3

3

)bkminusm(k) =

(k

m

)(k +m

m

)(B16)


Pk(u) =

ksumi=1

aiuiminus1 (B17)



bl(minus1)iminusl

(iminus l)= 0 (B18)




k = 2 ~a = a1 ak = 6 3

k = 3 ~a =1

3times 60 60 11

k = 4 ~a =5

6times 84 126 52 5

k = 5 ~a =1

30times 7560 15120 9870 2310 137

k = 6 ~a =7

10times 13720 3300 2960 1140 174 7 (B19)

which leads to

a1(k) =(2k)

(k)2

ak(k) = 2h(k) = 2

ksuml=1

1

l(B20)




ndash 31 ndash

JHEP03(2016)165


grals






infinsumli=0

summi

infinsumt=0

sump


(x1)Yl2minusm2

(x2)Yl3m3

(xprime3)Yl4m4



t+ 1

=x23x

24

x213x224

1

λlog

((1 + ρs)


)(B21)



C l1 l2 lm1m2 m1+m2

=

langl12

l22

mL12

mL22

∣∣∣∣∣ l2

mL1 +mL2

2

ranglangl12

l22

mR12

mR22

∣∣∣∣∣ l2

mR1 +mR22






Yn12

n12n12

(x1)Yn22

n22n22



(x4) = zn11 zn2

2 zn33 zn4

4 (B22)


zn11 zn2

2 zn33 zn4


note that





that zn11 zn2

2 zn33 zn4


Yn12

n12n1minus2

2

(x1)Yn22

n2minus22

n22

(x2)Yn32


2

(x3)Yn42

minusn4minus22

minusn42

(x4)

=radicn1n2n3n4z

n1minus11 w1z

n2minus12 w2z

n3minus13 w3z

n4minus14 w4 (B24)


coefficients

Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B25)

Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B26)

ndash 32 ndash

JHEP03(2016)165


Cn12n22n1+n2

2n12n2minus2

2n1+n2minus2

2

=

radicn2

n1 + n2(B27)

Cn12n22n1+n2minus2

2n12n2minus2

2n1+n2minus2

2

= minusradic

n1n1 + n2

(B28)



n2minus12 w2z

n3minus13 w3z

n4minus14 w4 is

radicn1n2n3n4

[Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2 + 1

+Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2

]=


(B29)




section 5






parth








prime4vminus) (C1)



(2ixprime3microx

prime3 middot

part


part


)eiKmiddotx

prime3vminus

= ix23part

partxmicro3

1

x23eiKmiddotx

prime3vminus (C2)


parth


part

partxmicro2

(1

x22h

)minus x23

part

partxmicro3

(1

x23h

)minus x24

part

partxmicro4

(1

x24h

)(C3)

ndash 33 ndash

JHEP03(2016)165


x22x23x

24

(part

partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)[h

x22x23x

24

]= 0 (C4)


x1micropart

partx1νminus x1ν

part

partx1micro

)h= F1(Mmicroνe









x22x23x

24

(x1micro

part

partx1νminus x1ν

part


part

partx2νminus x2ν

part


part

partx3νminus x3ν

part

partx3micro

+x4micropart

partx4νminus x4ν

part

partx4micro

)[h

x22x23x

24

]= 0 (C6)


consider(x1 middot

part

partx1+ 3

)h = F1(e






x22x23x

24

(x1 middot

part

partx1+ x2 middot

part

partx2+ x3 middot

part

partx3+ x4 middot

part

partx4+ 12

)[h

x22x23x

24

]= 0 (C8)


h

x22x23x

24

=A




we have




ndash 34 ndash

JHEP03(2016)165





infinsumnm=0

im(minus1)n




where







minus (C14)

we find


and

infinsumn=0

(minus1)n




1


(C16)



=infinsum

n1n3n4=0

1



slot we get

α(P middot x2) =1


ndash 35 ndash

JHEP03(2016)165




=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)


=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)

(n1 + n3 + n4)



=infinsumn1

(minus1)n1




timesinfinsumn3

(minus1)n3




timesinfinsumn4

(minus1)n4





=1


1


1


=x22x

33x

24






References














ndash 36 ndash

JHEP03(2016)165










(2003) 013 [hep-th0212208] [INSPIRE]




(2014) 302 [arXiv14036646] [INSPIRE]






















(2015) 006 [arXiv150208005] [INSPIRE]






ndash 37 ndash

JHEP03(2016)165



[INSPIRE]

















(2015) 29FT01 [arXiv150500963] [INSPIRE]








[INSPIRE]














ndash 38 ndash

JHEP03(2016)165















ndash 39 ndash

Introduction























JHEP03(2016)165











































ndash 4 ndash

JHEP03(2016)165















integrals














I(x1 x2 x3 x4) =

intd4x

2π21
















ndash 5 ndash

JHEP03(2016)165







prime4vminus (24)

To this we tensor

































ndash 6 ndash

JHEP03(2016)165


I(x1 x2 x3 x4) =1

2x213x224

Φ(s t) (26)

where

Φ(s t) =1

λ


(t

s

)ln

(1 + ρt

1 + ρs

)+π2

3

)(27)

and

ρ =2



s =x212x

234

x213x224

t =x214x

223

x213x224

(28)



λ sim u+ s

(minus1minus 2

u

)+ s2

(minus 2

u3minus 2

u2

)sim u+ s

(minus1minus 2

u

)ρ sim s2

(2

u5+

1

u4

)+

s


1


s(29)



6minus 1

2log2(minusz) (210)


Φ(s t) =1

λ


(st

)log


(1 + ρs)

)(211)


notation


F0(s u) =1

λ



(1 + ρs)

)F1(s u) = minus 1

λlog


(1 + ρs)









=f(x1 x

prime4)f(x2 x

prime3)


ndash 7 ndash

JHEP03(2016)165

where








in section 4









J3

∣∣∣minus 1

2

rang= minus1

2

∣∣∣minus 1

2

rangJminus

∣∣∣minus 1

2

rang= 0 (214)





JminusJ2+

∣∣∣minus1

2

rang= 0 (215)




(2)

minus 12

The quotient

space Vminus 12V

(2)

minus 12


sequence

0rarr V(2)

minus 12


2V

(2)

minus 12

rarr 0 (216)









ndash 8 ndash

JHEP03(2016)165










(p2)+ rarr 0 (220)
























in section 5


I1 =sumllimi

prod4i=1 Y

limi(x

primei)

|x1|2|x2|2|x3|2|x4|2

int |x1|0


4prodi=1

Y mili

(x) (31)

ndash 9 ndash

JHEP03(2016)165

where d3x = 12π2dS



write it as

Cm1m2m3m4

l1l2l3l4=

suml5m5m6

Cm1m2l5l1l2m5

1

(l5 + 1)Cm3m4l5l3l4m6


(32)

where

Y m1l1

Y m2l2

=suml5m5

Cm1m2l5l1l2m5

Y m5l5

(33)




or equivalently












I1 =sumlimi

prodi Y

limi(x

primei)

|x1|2|x2|2|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

|x1|4+l1+l2+l3+l44 + l1 + l2 + l3 + l4

(38)




ndash 10 ndash

JHEP03(2016)165


region is

I2 =sumlimi

Y l1m1

(x1)prodi=2 Y

limi(x

primei)

|x2|2|x3|2|x4|2

int |x2||x1|


l1(xprime)Y m2

l2(x)Y m3

l3(x)Y m4

l4(x)

=sumlimi

Y l1m1

(x1)prodi=2 Y

limi(x

primei)

|x2|2|x3|2|x4|2

int |x2||x1|


l1l2l3l4

=sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

|x2|2Y l3m3

(xprime3)

|x3|2Y l4m4

(xprime4)

|x4|2Cm1m2m3m4

l1l2l3l4

(minusl1 + l2 + l3 + l4 + 2)


1

)(39)





IS21 =sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

|x2|2Y l3m3

(xprime3)

|x3|2Y l4m4

(xprime4)

|x4|2Cm1m2m3m4

l1l2l3l4

(minusl1 + l2 + l3 + l4 + 2)


1

)IS22 =

sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

|x2|2Y l3m3

(xprime3)

|x3|2Y l4m4

(xprime4)

|x4|2Cm1m2m3m4

l1l2l3l4

(minusl1 + l2 + l3 + l4 + 2)


)(310)




primemicro3 x



I2 = IS21 + IR21 (311)


I3 =sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

int |x3||x2|


=sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4log

(|x3||x2|

)δ(l1 + l2 l3 + l4)

+sum

limil 6=0

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2

(|x3|l minus |x2|l

l


(312)

ndash 11 ndash

JHEP03(2016)165


IL32 =minussumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4log (|x2|) δ(l1 + l2 l3 + l4)

IS32 =sum

limil 6=0

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

(minus|x2|l

l


IL33 =sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4log (|x3|) δ(l1 + l2 l3 + l4)

IS33 =sum

limil 6=0

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

(|x3|l

l


(313)




I3 = IL32 + IS32 + IL33 + IS33 (314)



we know that

|x2| = max(r1 r2) =1

2(r1 + r2 + |r1 minus r2|)

|x3| = min(r3 r4) =1

2(r3 + r4 minus |r3 minus r4|) (315)



)(316)





I4 =sumlimi

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)


int |x4||x3|


=sum

limille0

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)

|x4|2


l minus 2



I43 =sum

limille0

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)

|x4|2

(minus|x3|lminus2

l minus 2


I44 =sum

limille0

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)

|x4|2

(|x4|lminus2

l minus 2

)(318)

ndash 12 ndash

JHEP03(2016)165




I4 = I43 + I44 (319)


I5 =sumlimi

Cm1m2m3m4

l1l2l3l4

prodi

Y limi(xi)

int infin|x4|


=sumlimil

Cm1m2m3m4

l1l2l3l4

prodi

Y limi(xi)

(|x4|minus4minusl

4 + l

)δ(l l1 + l2 + l3 + l4) (320)






C2 = ∆(∆minus 4) + l(l + 2) = minus3 (321)










and Libine [29]


result is

log(r3r2

)sumlimi

(Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

)δ(l1 + l2 l3 + l4) (41)


Φ(s t) = F0(s t) + log(s)F1(s t) (42)



ndash 13 ndash

JHEP03(2016)165





)spart

parts

(spartH

parts

)minus(


(tpartH

partt

)+ 4s

partH

parts(44)


C2 sΦ = 2 sΦ C2 sF1 = 2 sF1 (45)



for the integral is

1

2x213x224

F1(s t) (46)



plies that

2sumlimi

(Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

)δ(l1 + l2 l3 + l4) =

1

x213x224

F1(s t)

(47)




V+ otimes V+ = A[200] +

infinoplusk=1

D[k+2 k2 k2] (48)








P++++ defined by




ndash 14 ndash

JHEP03(2016)165



















1

2x213x224



prime24 P(x1 x2 x










ndash 15 ndash

JHEP03(2016)165





m1(x)Y l2

m2(x)Y l5

m5(x) (416)


(x1)Yl2m2


Y l1m1

(x1)otimes Y l2m2


(x)Y l2m2

(x)

Y l3m3

(xprime3)

|x3|2otimesY l4m4

(xprime4)

|x4|2rarr

Y l3m3

(xprime)

|x|2otimesY l4m4

(xprime)

|x|2(417)















m(K)Y ml (xprime)

sumlimi

(minusi)l12l1 l1

Y l1m1


Y l2m2

(x2)(i)l3

2l3 l3Y l3m3

(xprime3)(i)l4

2l4 l4Y l4m4

(xprime4)

P++minusminus

(Y m1l1




(K)vminus)

=sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

P++minusminus

(Y m1l1

(P )

2l1 l1v+ otimes

Y m2l2

(P )

2l2 l2v+ otimes

Y m3l3

(K)

2l3 l3vminus otimes

Y m4l4

(K)

2l4 l4vminus)

(418)








ndash 16 ndash

JHEP03(2016)165






F1 =1

λlog

((1 + ρs)


)(419)



sk

u2k+1[Qk(u) log(1 + u)]+ (420)



Qk(u) =

ksuml=0

bl(k)ul (421)

where

bl(k) =

(k

k minus l

)(2k + l

k minus l

)(422)








equivariant maps


I = I1 + I2 + I3 + I4 + I5



I(1) = (IS11 + IS21)

I(2) = (IS22 + IS32 + IL32)

I(3) = (IL33 + IS33 + IS43)

I(4) = (IS45 + IS55) (52)

ndash 17 ndash

JHEP03(2016)165




I = I(1) + I(2) + I(3) + I(4) (53)



jI(i) = minus2δij

prodk 6=j

1

(xk minus xi)2(54)









I(1) =sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

r22

Y l3m3

(xprime3)

r23

Y l4m4

(xprime4)



(l1 + l2 + l3 + l4 + 4)(minusl1 + l2 + l3 + l4 + 2)(55)




part2


(r21)A = 4A(A+ 1)(r21)Aminus1

part


part

partxmicro1Y l1m1

(x1)part


21)Aminus1Y l1

m1(x1) (56)


part2


I(1) =sumlimi


r22r23r

24

Y l1m1

(x1)Yl2m2

(xprime2)Yl3m3

(xprime3)Yl4m4


(57)


1


1

r22r23r

24

sumlimi

Y l2m2

(x1)Yl3m3

(x1)Yl4m4

(x1)Yl2m2

(xprime2)Yl3m3

(xprime3)Yl4m4

(xprime4)

=(l1 + 1)rl2+l3+l41

r22r23r

24

sumlimi

Cm1m2m3m4l1l2l3l4

Y l1m1

(x)Y l2m2

(xprime2)Yl3m3

(xprime3)Yl4m4

(xprime4) (58)

ndash 18 ndash

JHEP03(2016)165














I(2) =sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


1

l2+l3+l4minusl1+2

δ(l1 + l2 l3 + l4)

r23r24

+sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


1

r23r24

times



]


part

partxmicro2Y l2m2

(x2)part


22)Aminus1Y l2

m2(x2)

part

partxmicro2logr2 =

xmicro2r22

part

partxmicro2

part

partxmicro2logr2 =

2

r22(59)

We find

part2




r23r24

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


(510)


minus 2







ndash 19 ndash

JHEP03(2016)165


equivariant map


such that



x22x23x

24








23x

24


prime2)f(x1 x

prime3)f(x1 x

prime4)

(515)



1


(516)







part2



prime2)f(x1 x

prime3)f(x1 x

prime4) (518)


24I

(1)(x1 x2 x3 x4)


x22x23x

24 I(1)(x1 x2 x3 x4)rarr F1(e


While x23x24I








ndash 20 ndash

JHEP03(2016)165

implies that (part

partxmicro1+ x22

part

partxmicro2

1

x22+ x23

part

partxmicro3

1

x23+ x24

part

partxmicro4

1

x24

)F1 = 0 (521)


24I



24(I

(1)+I(2)+I(3)+I(4)) = x22x23x



partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)I = 0 (522)





23x

24I(x1 x2 x3 x4) =

x22x23x

24

x213x224

Φ(s t)

(523)



with


minus)=xprime21 x

prime22 xprime23


The x21x22x

23I













〈φ(x1)φ3(x2)φ(x3)φ(x4)〉 (528)




ndash 21 ndash

JHEP03(2016)165



twisted co-product










23x

24



















prime22 xprime23






ndash 22 ndash

JHEP03(2016)165





I2 =

intd4x5d

4x6xminus256

x215x225x

245x

226x

246x

236

(61)






















ndash 23 ndash

JHEP03(2016)165
































ndash 24 ndash

JHEP03(2016)165































Acknowledgments








this work was done

ndash 25 ndash

JHEP03(2016)165








1

2π2

intdS3Y l1

m1(x)Y l2

m2(x) = gm1m2

δl1l2

(l1 + 1)(A1)




gm1m2 = δm1m2


m2(x) = Y l

m1(x)

Y ml (x) = Y l

m(x) (A2)






(mLmR)

gm1m2 = δm1minusm2

Y ml (x) = Y l

minusm(x) (A3)



ables are

z = x1 + ix2 harr(

1

2

1

2


(minus1

2minus1

2

)w = x3 + ix4 harr

(1

2minus1

2


(minus1

21

2

)(A4)

We find that

Y l(l2l2) = (x1 + ix2)

l = zl (A5)

has normalization

1

2π2

intdS3(Y l

( l2 l2))lowast Y l

( l2 l2)

=1

(l + 1)(A6)


ndash 26 ndash

JHEP03(2016)165




(minusi)l

2llY lm(x)Y m

l (P )v+ (A7)


η(Y l1m1

(P )v+ Y m2l2

(K)vminus)

= δl1l2δm1m222l1(l1)2 (A8)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)= δl1l2δm1m2 (A9)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)=

1

2π2

intS3

Y m2l2

(xprime)


m1(x)) (A10)





Y lm(xprime)

xprime2


m(x) (A12)


η = f η


1


summ

1

y2Y lm(x)Y m


=infinsuml=0

summ

1

x2Y lm(xprime)Y m

l (y) if |x| gt |y| (A13)






ndash 27 ndash

JHEP03(2016)165


C2 = minus1

2LMNL


2MpqMqp (A14)






(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)eminusiP middotxv+

=

(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)ipart





Mpq = xppart

partxqminus xq

part

partxp(A16)


1

2MpqMqp(r

AY lm(x)) =

rA

2MpqMqpY


m(x) (A17)





m(x) (A19)

Thus we have

C2 rAY l




m(x) (A21)





F1 =1

λlog

((1 + ρs)


)(B1)

ndash 28 ndash

JHEP03(2016)165


of s to obtain

F1 =log(1+u)

u+

s


s2

u5(3u(2+u)minus 6 + 6u+ u2) log(1+u)

)+s3


)(B2)


F1 =

(minus1 +

u

2minus u2

3+u3


)+ s

(minus1

6+u

6minus 3u2


)+ s2

(minus1

30+

u

20minus 2u2


)+s3

(minus1

140+

u

70minus 5u2



The term at sk is

sk









in Pk(u) is

(2k)

(k)2(B6)


Qk(u) =ksuml=0

blul (B7)


ksuml=0

bl(minus1)l

(iminus l)= 0 (B8)




b1 =5b02 b2 =

20b09 b3 =

5b06 b4 =

5b042 b5 =

b0252

(B9)

ndash 29 ndash

JHEP03(2016)165



ksuml=0

MIlbl = 0 (B10)

where

MIl =(minus1)l




MIlbl (B12)



bl =sumI




bIb0 = 1 16bIb0 = 32 35 120bIb0 = 2 97 27 170

bIb0 =

5

2

20

95

6

5

42

1

252

bIb0 =

3

75

22

20

11

5

11

1

22

1

924

bIb0 =

7

2

63

13175

52

175

143

63

286

7

429

1

3432

(B14)



b1(k)b0 = k2

bk(k)b0 =k2

(2k)

bkminus1(k)b0 =k2

(2k)k(k + 1)

bkminus2(k)b0 =k2

(2k)

(k minus 1)k(k + 1)(k + 2)

4

bkminus3(k)b0 =k2

(2k)

(k

3

)(k + 3

3

)bkminusm(k)b0 =

k2

(2k)

(k

m

)(k +m

m

)(B15)

ndash 30 ndash

JHEP03(2016)165


so that

we have

b1(k) = k2(2k)

k2

bk(k) = 1



4

bkminus3(k) =

(k

3

)(k + 3

3

)bkminusm(k) =

(k

m

)(k +m

m

)(B16)


Pk(u) =

ksumi=1

aiuiminus1 (B17)



bl(minus1)iminusl

(iminus l)= 0 (B18)




k = 2 ~a = a1 ak = 6 3

k = 3 ~a =1

3times 60 60 11

k = 4 ~a =5

6times 84 126 52 5

k = 5 ~a =1

30times 7560 15120 9870 2310 137

k = 6 ~a =7

10times 13720 3300 2960 1140 174 7 (B19)

which leads to

a1(k) =(2k)

(k)2

ak(k) = 2h(k) = 2

ksuml=1

1

l(B20)




ndash 31 ndash

JHEP03(2016)165


grals






infinsumli=0

summi

infinsumt=0

sump


(x1)Yl2minusm2

(x2)Yl3m3

(xprime3)Yl4m4



t+ 1

=x23x

24

x213x224

1

λlog

((1 + ρs)


)(B21)



C l1 l2 lm1m2 m1+m2

=

langl12

l22

mL12

mL22

∣∣∣∣∣ l2

mL1 +mL2

2

ranglangl12

l22

mR12

mR22

∣∣∣∣∣ l2

mR1 +mR22






Yn12

n12n12

(x1)Yn22

n22n22



(x4) = zn11 zn2

2 zn33 zn4

4 (B22)


zn11 zn2

2 zn33 zn4


note that





that zn11 zn2

2 zn33 zn4


Yn12

n12n1minus2

2

(x1)Yn22

n2minus22

n22

(x2)Yn32


2

(x3)Yn42

minusn4minus22

minusn42

(x4)

=radicn1n2n3n4z

n1minus11 w1z

n2minus12 w2z

n3minus13 w3z

n4minus14 w4 (B24)


coefficients

Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B25)

Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B26)

ndash 32 ndash

JHEP03(2016)165


Cn12n22n1+n2

2n12n2minus2

2n1+n2minus2

2

=

radicn2

n1 + n2(B27)

Cn12n22n1+n2minus2

2n12n2minus2

2n1+n2minus2

2

= minusradic

n1n1 + n2

(B28)



n2minus12 w2z

n3minus13 w3z

n4minus14 w4 is

radicn1n2n3n4

[Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2 + 1

+Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2

]=


(B29)




section 5






parth








prime4vminus) (C1)



(2ixprime3microx

prime3 middot

part


part


)eiKmiddotx

prime3vminus

= ix23part

partxmicro3

1

x23eiKmiddotx

prime3vminus (C2)


parth


part

partxmicro2

(1

x22h

)minus x23

part

partxmicro3

(1

x23h

)minus x24

part

partxmicro4

(1

x24h

)(C3)

ndash 33 ndash

JHEP03(2016)165


x22x23x

24

(part

partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)[h

x22x23x

24

]= 0 (C4)


x1micropart

partx1νminus x1ν

part

partx1micro

)h= F1(Mmicroνe









x22x23x

24

(x1micro

part

partx1νminus x1ν

part


part

partx2νminus x2ν

part


part

partx3νminus x3ν

part

partx3micro

+x4micropart

partx4νminus x4ν

part

partx4micro

)[h

x22x23x

24

]= 0 (C6)


consider(x1 middot

part

partx1+ 3

)h = F1(e






x22x23x

24

(x1 middot

part

partx1+ x2 middot

part

partx2+ x3 middot

part

partx3+ x4 middot

part

partx4+ 12

)[h

x22x23x

24

]= 0 (C8)


h

x22x23x

24

=A




we have




ndash 34 ndash

JHEP03(2016)165





infinsumnm=0

im(minus1)n




where







minus (C14)

we find


and

infinsumn=0

(minus1)n




1


(C16)



=infinsum

n1n3n4=0

1



slot we get

α(P middot x2) =1


ndash 35 ndash

JHEP03(2016)165




=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)


=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)

(n1 + n3 + n4)



=infinsumn1

(minus1)n1




timesinfinsumn3

(minus1)n3




timesinfinsumn4

(minus1)n4





=1


1


1


=x22x

33x

24






References














ndash 36 ndash

JHEP03(2016)165










(2003) 013 [hep-th0212208] [INSPIRE]




(2014) 302 [arXiv14036646] [INSPIRE]






















(2015) 006 [arXiv150208005] [INSPIRE]






ndash 37 ndash

JHEP03(2016)165



[INSPIRE]

















(2015) 29FT01 [arXiv150500963] [INSPIRE]








[INSPIRE]














ndash 38 ndash

JHEP03(2016)165















ndash 39 ndash

Introduction























JHEP03(2016)165















integrals














I(x1 x2 x3 x4) =

intd4x

2π21
















ndash 5 ndash

JHEP03(2016)165







prime4vminus (24)

To this we tensor

































ndash 6 ndash

JHEP03(2016)165


I(x1 x2 x3 x4) =1

2x213x224

Φ(s t) (26)

where

Φ(s t) =1

λ


(t

s

)ln

(1 + ρt

1 + ρs

)+π2

3

)(27)

and

ρ =2



s =x212x

234

x213x224

t =x214x

223

x213x224

(28)



λ sim u+ s

(minus1minus 2

u

)+ s2

(minus 2

u3minus 2

u2

)sim u+ s

(minus1minus 2

u

)ρ sim s2

(2

u5+

1

u4

)+

s


1


s(29)



6minus 1

2log2(minusz) (210)


Φ(s t) =1

λ


(st

)log


(1 + ρs)

)(211)


notation


F0(s u) =1

λ



(1 + ρs)

)F1(s u) = minus 1

λlog


(1 + ρs)









=f(x1 x

prime4)f(x2 x

prime3)


ndash 7 ndash

JHEP03(2016)165

where








in section 4









J3

∣∣∣minus 1

2

rang= minus1

2

∣∣∣minus 1

2

rangJminus

∣∣∣minus 1

2

rang= 0 (214)





JminusJ2+

∣∣∣minus1

2

rang= 0 (215)




(2)

minus 12

The quotient

space Vminus 12V

(2)

minus 12


sequence

0rarr V(2)

minus 12


2V

(2)

minus 12

rarr 0 (216)









ndash 8 ndash

JHEP03(2016)165










(p2)+ rarr 0 (220)
























in section 5


I1 =sumllimi

prod4i=1 Y

limi(x

primei)

|x1|2|x2|2|x3|2|x4|2

int |x1|0


4prodi=1

Y mili

(x) (31)

ndash 9 ndash

JHEP03(2016)165

where d3x = 12π2dS



write it as

Cm1m2m3m4

l1l2l3l4=

suml5m5m6

Cm1m2l5l1l2m5

1

(l5 + 1)Cm3m4l5l3l4m6


(32)

where

Y m1l1

Y m2l2

=suml5m5

Cm1m2l5l1l2m5

Y m5l5

(33)




or equivalently












I1 =sumlimi

prodi Y

limi(x

primei)

|x1|2|x2|2|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

|x1|4+l1+l2+l3+l44 + l1 + l2 + l3 + l4

(38)




ndash 10 ndash

JHEP03(2016)165


region is

I2 =sumlimi

Y l1m1

(x1)prodi=2 Y

limi(x

primei)

|x2|2|x3|2|x4|2

int |x2||x1|


l1(xprime)Y m2

l2(x)Y m3

l3(x)Y m4

l4(x)

=sumlimi

Y l1m1

(x1)prodi=2 Y

limi(x

primei)

|x2|2|x3|2|x4|2

int |x2||x1|


l1l2l3l4

=sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

|x2|2Y l3m3

(xprime3)

|x3|2Y l4m4

(xprime4)

|x4|2Cm1m2m3m4

l1l2l3l4

(minusl1 + l2 + l3 + l4 + 2)


1

)(39)





IS21 =sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

|x2|2Y l3m3

(xprime3)

|x3|2Y l4m4

(xprime4)

|x4|2Cm1m2m3m4

l1l2l3l4

(minusl1 + l2 + l3 + l4 + 2)


1

)IS22 =

sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

|x2|2Y l3m3

(xprime3)

|x3|2Y l4m4

(xprime4)

|x4|2Cm1m2m3m4

l1l2l3l4

(minusl1 + l2 + l3 + l4 + 2)


)(310)




primemicro3 x



I2 = IS21 + IR21 (311)


I3 =sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

int |x3||x2|


=sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4log

(|x3||x2|

)δ(l1 + l2 l3 + l4)

+sum

limil 6=0

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2

(|x3|l minus |x2|l

l


(312)

ndash 11 ndash

JHEP03(2016)165


IL32 =minussumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4log (|x2|) δ(l1 + l2 l3 + l4)

IS32 =sum

limil 6=0

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

(minus|x2|l

l


IL33 =sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4log (|x3|) δ(l1 + l2 l3 + l4)

IS33 =sum

limil 6=0

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

(|x3|l

l


(313)




I3 = IL32 + IS32 + IL33 + IS33 (314)



we know that

|x2| = max(r1 r2) =1

2(r1 + r2 + |r1 minus r2|)

|x3| = min(r3 r4) =1

2(r3 + r4 minus |r3 minus r4|) (315)



)(316)





I4 =sumlimi

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)


int |x4||x3|


=sum

limille0

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)

|x4|2


l minus 2



I43 =sum

limille0

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)

|x4|2

(minus|x3|lminus2

l minus 2


I44 =sum

limille0

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)

|x4|2

(|x4|lminus2

l minus 2

)(318)

ndash 12 ndash

JHEP03(2016)165




I4 = I43 + I44 (319)


I5 =sumlimi

Cm1m2m3m4

l1l2l3l4

prodi

Y limi(xi)

int infin|x4|


=sumlimil

Cm1m2m3m4

l1l2l3l4

prodi

Y limi(xi)

(|x4|minus4minusl

4 + l

)δ(l l1 + l2 + l3 + l4) (320)






C2 = ∆(∆minus 4) + l(l + 2) = minus3 (321)










and Libine [29]


result is

log(r3r2

)sumlimi

(Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

)δ(l1 + l2 l3 + l4) (41)


Φ(s t) = F0(s t) + log(s)F1(s t) (42)



ndash 13 ndash

JHEP03(2016)165





)spart

parts

(spartH

parts

)minus(


(tpartH

partt

)+ 4s

partH

parts(44)


C2 sΦ = 2 sΦ C2 sF1 = 2 sF1 (45)



for the integral is

1

2x213x224

F1(s t) (46)



plies that

2sumlimi

(Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

)δ(l1 + l2 l3 + l4) =

1

x213x224

F1(s t)

(47)




V+ otimes V+ = A[200] +

infinoplusk=1

D[k+2 k2 k2] (48)








P++++ defined by




ndash 14 ndash

JHEP03(2016)165



















1

2x213x224



prime24 P(x1 x2 x










ndash 15 ndash

JHEP03(2016)165





m1(x)Y l2

m2(x)Y l5

m5(x) (416)


(x1)Yl2m2


Y l1m1

(x1)otimes Y l2m2


(x)Y l2m2

(x)

Y l3m3

(xprime3)

|x3|2otimesY l4m4

(xprime4)

|x4|2rarr

Y l3m3

(xprime)

|x|2otimesY l4m4

(xprime)

|x|2(417)















m(K)Y ml (xprime)

sumlimi

(minusi)l12l1 l1

Y l1m1


Y l2m2

(x2)(i)l3

2l3 l3Y l3m3

(xprime3)(i)l4

2l4 l4Y l4m4

(xprime4)

P++minusminus

(Y m1l1




(K)vminus)

=sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

P++minusminus

(Y m1l1

(P )

2l1 l1v+ otimes

Y m2l2

(P )

2l2 l2v+ otimes

Y m3l3

(K)

2l3 l3vminus otimes

Y m4l4

(K)

2l4 l4vminus)

(418)








ndash 16 ndash

JHEP03(2016)165






F1 =1

λlog

((1 + ρs)


)(419)



sk

u2k+1[Qk(u) log(1 + u)]+ (420)



Qk(u) =

ksuml=0

bl(k)ul (421)

where

bl(k) =

(k

k minus l

)(2k + l

k minus l

)(422)








equivariant maps


I = I1 + I2 + I3 + I4 + I5



I(1) = (IS11 + IS21)

I(2) = (IS22 + IS32 + IL32)

I(3) = (IL33 + IS33 + IS43)

I(4) = (IS45 + IS55) (52)

ndash 17 ndash

JHEP03(2016)165




I = I(1) + I(2) + I(3) + I(4) (53)



jI(i) = minus2δij

prodk 6=j

1

(xk minus xi)2(54)









I(1) =sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

r22

Y l3m3

(xprime3)

r23

Y l4m4

(xprime4)



(l1 + l2 + l3 + l4 + 4)(minusl1 + l2 + l3 + l4 + 2)(55)




part2


(r21)A = 4A(A+ 1)(r21)Aminus1

part


part

partxmicro1Y l1m1

(x1)part


21)Aminus1Y l1

m1(x1) (56)


part2


I(1) =sumlimi


r22r23r

24

Y l1m1

(x1)Yl2m2

(xprime2)Yl3m3

(xprime3)Yl4m4


(57)


1


1

r22r23r

24

sumlimi

Y l2m2

(x1)Yl3m3

(x1)Yl4m4

(x1)Yl2m2

(xprime2)Yl3m3

(xprime3)Yl4m4

(xprime4)

=(l1 + 1)rl2+l3+l41

r22r23r

24

sumlimi

Cm1m2m3m4l1l2l3l4

Y l1m1

(x)Y l2m2

(xprime2)Yl3m3

(xprime3)Yl4m4

(xprime4) (58)

ndash 18 ndash

JHEP03(2016)165














I(2) =sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


1

l2+l3+l4minusl1+2

δ(l1 + l2 l3 + l4)

r23r24

+sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


1

r23r24

times



]


part

partxmicro2Y l2m2

(x2)part


22)Aminus1Y l2

m2(x2)

part

partxmicro2logr2 =

xmicro2r22

part

partxmicro2

part

partxmicro2logr2 =

2

r22(59)

We find

part2




r23r24

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


(510)


minus 2







ndash 19 ndash

JHEP03(2016)165


equivariant map


such that



x22x23x

24








23x

24


prime2)f(x1 x

prime3)f(x1 x

prime4)

(515)



1


(516)







part2



prime2)f(x1 x

prime3)f(x1 x

prime4) (518)


24I

(1)(x1 x2 x3 x4)


x22x23x

24 I(1)(x1 x2 x3 x4)rarr F1(e


While x23x24I








ndash 20 ndash

JHEP03(2016)165

implies that (part

partxmicro1+ x22

part

partxmicro2

1

x22+ x23

part

partxmicro3

1

x23+ x24

part

partxmicro4

1

x24

)F1 = 0 (521)


24I



24(I

(1)+I(2)+I(3)+I(4)) = x22x23x



partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)I = 0 (522)





23x

24I(x1 x2 x3 x4) =

x22x23x

24

x213x224

Φ(s t)

(523)



with


minus)=xprime21 x

prime22 xprime23


The x21x22x

23I













〈φ(x1)φ3(x2)φ(x3)φ(x4)〉 (528)




ndash 21 ndash

JHEP03(2016)165



twisted co-product










23x

24



















prime22 xprime23






ndash 22 ndash

JHEP03(2016)165





I2 =

intd4x5d

4x6xminus256

x215x225x

245x

226x

246x

236

(61)






















ndash 23 ndash

JHEP03(2016)165
































ndash 24 ndash

JHEP03(2016)165































Acknowledgments








this work was done

ndash 25 ndash

JHEP03(2016)165








1

2π2

intdS3Y l1

m1(x)Y l2

m2(x) = gm1m2

δl1l2

(l1 + 1)(A1)




gm1m2 = δm1m2


m2(x) = Y l

m1(x)

Y ml (x) = Y l

m(x) (A2)






(mLmR)

gm1m2 = δm1minusm2

Y ml (x) = Y l

minusm(x) (A3)



ables are

z = x1 + ix2 harr(

1

2

1

2


(minus1

2minus1

2

)w = x3 + ix4 harr

(1

2minus1

2


(minus1

21

2

)(A4)

We find that

Y l(l2l2) = (x1 + ix2)

l = zl (A5)

has normalization

1

2π2

intdS3(Y l

( l2 l2))lowast Y l

( l2 l2)

=1

(l + 1)(A6)


ndash 26 ndash

JHEP03(2016)165




(minusi)l

2llY lm(x)Y m

l (P )v+ (A7)


η(Y l1m1

(P )v+ Y m2l2

(K)vminus)

= δl1l2δm1m222l1(l1)2 (A8)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)= δl1l2δm1m2 (A9)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)=

1

2π2

intS3

Y m2l2

(xprime)


m1(x)) (A10)





Y lm(xprime)

xprime2


m(x) (A12)


η = f η


1


summ

1

y2Y lm(x)Y m


=infinsuml=0

summ

1

x2Y lm(xprime)Y m

l (y) if |x| gt |y| (A13)






ndash 27 ndash

JHEP03(2016)165


C2 = minus1

2LMNL


2MpqMqp (A14)






(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)eminusiP middotxv+

=

(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)ipart





Mpq = xppart

partxqminus xq

part

partxp(A16)


1

2MpqMqp(r

AY lm(x)) =

rA

2MpqMqpY


m(x) (A17)





m(x) (A19)

Thus we have

C2 rAY l




m(x) (A21)





F1 =1

λlog

((1 + ρs)


)(B1)

ndash 28 ndash

JHEP03(2016)165


of s to obtain

F1 =log(1+u)

u+

s


s2

u5(3u(2+u)minus 6 + 6u+ u2) log(1+u)

)+s3


)(B2)


F1 =

(minus1 +

u

2minus u2

3+u3


)+ s

(minus1

6+u

6minus 3u2


)+ s2

(minus1

30+

u

20minus 2u2


)+s3

(minus1

140+

u

70minus 5u2



The term at sk is

sk









in Pk(u) is

(2k)

(k)2(B6)


Qk(u) =ksuml=0

blul (B7)


ksuml=0

bl(minus1)l

(iminus l)= 0 (B8)




b1 =5b02 b2 =

20b09 b3 =

5b06 b4 =

5b042 b5 =

b0252

(B9)

ndash 29 ndash

JHEP03(2016)165



ksuml=0

MIlbl = 0 (B10)

where

MIl =(minus1)l




MIlbl (B12)



bl =sumI




bIb0 = 1 16bIb0 = 32 35 120bIb0 = 2 97 27 170

bIb0 =

5

2

20

95

6

5

42

1

252

bIb0 =

3

75

22

20

11

5

11

1

22

1

924

bIb0 =

7

2

63

13175

52

175

143

63

286

7

429

1

3432

(B14)



b1(k)b0 = k2

bk(k)b0 =k2

(2k)

bkminus1(k)b0 =k2

(2k)k(k + 1)

bkminus2(k)b0 =k2

(2k)

(k minus 1)k(k + 1)(k + 2)

4

bkminus3(k)b0 =k2

(2k)

(k

3

)(k + 3

3

)bkminusm(k)b0 =

k2

(2k)

(k

m

)(k +m

m

)(B15)

ndash 30 ndash

JHEP03(2016)165


so that

we have

b1(k) = k2(2k)

k2

bk(k) = 1



4

bkminus3(k) =

(k

3

)(k + 3

3

)bkminusm(k) =

(k

m

)(k +m

m

)(B16)


Pk(u) =

ksumi=1

aiuiminus1 (B17)



bl(minus1)iminusl

(iminus l)= 0 (B18)




k = 2 ~a = a1 ak = 6 3

k = 3 ~a =1

3times 60 60 11

k = 4 ~a =5

6times 84 126 52 5

k = 5 ~a =1

30times 7560 15120 9870 2310 137

k = 6 ~a =7

10times 13720 3300 2960 1140 174 7 (B19)

which leads to

a1(k) =(2k)

(k)2

ak(k) = 2h(k) = 2

ksuml=1

1

l(B20)




ndash 31 ndash

JHEP03(2016)165


grals






infinsumli=0

summi

infinsumt=0

sump


(x1)Yl2minusm2

(x2)Yl3m3

(xprime3)Yl4m4



t+ 1

=x23x

24

x213x224

1

λlog

((1 + ρs)


)(B21)



C l1 l2 lm1m2 m1+m2

=

langl12

l22

mL12

mL22

∣∣∣∣∣ l2

mL1 +mL2

2

ranglangl12

l22

mR12

mR22

∣∣∣∣∣ l2

mR1 +mR22






Yn12

n12n12

(x1)Yn22

n22n22



(x4) = zn11 zn2

2 zn33 zn4

4 (B22)


zn11 zn2

2 zn33 zn4


note that





that zn11 zn2

2 zn33 zn4


Yn12

n12n1minus2

2

(x1)Yn22

n2minus22

n22

(x2)Yn32


2

(x3)Yn42

minusn4minus22

minusn42

(x4)

=radicn1n2n3n4z

n1minus11 w1z

n2minus12 w2z

n3minus13 w3z

n4minus14 w4 (B24)


coefficients

Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B25)

Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B26)

ndash 32 ndash

JHEP03(2016)165


Cn12n22n1+n2

2n12n2minus2

2n1+n2minus2

2

=

radicn2

n1 + n2(B27)

Cn12n22n1+n2minus2

2n12n2minus2

2n1+n2minus2

2

= minusradic

n1n1 + n2

(B28)



n2minus12 w2z

n3minus13 w3z

n4minus14 w4 is

radicn1n2n3n4

[Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2 + 1

+Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2

]=


(B29)




section 5






parth








prime4vminus) (C1)



(2ixprime3microx

prime3 middot

part


part


)eiKmiddotx

prime3vminus

= ix23part

partxmicro3

1

x23eiKmiddotx

prime3vminus (C2)


parth


part

partxmicro2

(1

x22h

)minus x23

part

partxmicro3

(1

x23h

)minus x24

part

partxmicro4

(1

x24h

)(C3)

ndash 33 ndash

JHEP03(2016)165


x22x23x

24

(part

partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)[h

x22x23x

24

]= 0 (C4)


x1micropart

partx1νminus x1ν

part

partx1micro

)h= F1(Mmicroνe









x22x23x

24

(x1micro

part

partx1νminus x1ν

part


part

partx2νminus x2ν

part


part

partx3νminus x3ν

part

partx3micro

+x4micropart

partx4νminus x4ν

part

partx4micro

)[h

x22x23x

24

]= 0 (C6)


consider(x1 middot

part

partx1+ 3

)h = F1(e






x22x23x

24

(x1 middot

part

partx1+ x2 middot

part

partx2+ x3 middot

part

partx3+ x4 middot

part

partx4+ 12

)[h

x22x23x

24

]= 0 (C8)


h

x22x23x

24

=A




we have




ndash 34 ndash

JHEP03(2016)165





infinsumnm=0

im(minus1)n




where







minus (C14)

we find


and

infinsumn=0

(minus1)n




1


(C16)



=infinsum

n1n3n4=0

1



slot we get

α(P middot x2) =1


ndash 35 ndash

JHEP03(2016)165




=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)


=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)

(n1 + n3 + n4)



=infinsumn1

(minus1)n1




timesinfinsumn3

(minus1)n3




timesinfinsumn4

(minus1)n4





=1


1


1


=x22x

33x

24






References














ndash 36 ndash

JHEP03(2016)165










(2003) 013 [hep-th0212208] [INSPIRE]




(2014) 302 [arXiv14036646] [INSPIRE]






















(2015) 006 [arXiv150208005] [INSPIRE]






ndash 37 ndash

JHEP03(2016)165



[INSPIRE]

















(2015) 29FT01 [arXiv150500963] [INSPIRE]








[INSPIRE]














ndash 38 ndash

JHEP03(2016)165















ndash 39 ndash

Introduction























JHEP03(2016)165







prime4vminus (24)

To this we tensor

































ndash 6 ndash

JHEP03(2016)165


I(x1 x2 x3 x4) =1

2x213x224

Φ(s t) (26)

where

Φ(s t) =1

λ


(t

s

)ln

(1 + ρt

1 + ρs

)+π2

3

)(27)

and

ρ =2



s =x212x

234

x213x224

t =x214x

223

x213x224

(28)



λ sim u+ s

(minus1minus 2

u

)+ s2

(minus 2

u3minus 2

u2

)sim u+ s

(minus1minus 2

u

)ρ sim s2

(2

u5+

1

u4

)+

s


1


s(29)



6minus 1

2log2(minusz) (210)


Φ(s t) =1

λ


(st

)log


(1 + ρs)

)(211)


notation


F0(s u) =1

λ



(1 + ρs)

)F1(s u) = minus 1

λlog


(1 + ρs)









=f(x1 x

prime4)f(x2 x

prime3)


ndash 7 ndash

JHEP03(2016)165

where








in section 4









J3

∣∣∣minus 1

2

rang= minus1

2

∣∣∣minus 1

2

rangJminus

∣∣∣minus 1

2

rang= 0 (214)





JminusJ2+

∣∣∣minus1

2

rang= 0 (215)




(2)

minus 12

The quotient

space Vminus 12V

(2)

minus 12


sequence

0rarr V(2)

minus 12


2V

(2)

minus 12

rarr 0 (216)









ndash 8 ndash

JHEP03(2016)165










(p2)+ rarr 0 (220)
























in section 5


I1 =sumllimi

prod4i=1 Y

limi(x

primei)

|x1|2|x2|2|x3|2|x4|2

int |x1|0


4prodi=1

Y mili

(x) (31)

ndash 9 ndash

JHEP03(2016)165

where d3x = 12π2dS



write it as

Cm1m2m3m4

l1l2l3l4=

suml5m5m6

Cm1m2l5l1l2m5

1

(l5 + 1)Cm3m4l5l3l4m6


(32)

where

Y m1l1

Y m2l2

=suml5m5

Cm1m2l5l1l2m5

Y m5l5

(33)




or equivalently












I1 =sumlimi

prodi Y

limi(x

primei)

|x1|2|x2|2|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

|x1|4+l1+l2+l3+l44 + l1 + l2 + l3 + l4

(38)




ndash 10 ndash

JHEP03(2016)165


region is

I2 =sumlimi

Y l1m1

(x1)prodi=2 Y

limi(x

primei)

|x2|2|x3|2|x4|2

int |x2||x1|


l1(xprime)Y m2

l2(x)Y m3

l3(x)Y m4

l4(x)

=sumlimi

Y l1m1

(x1)prodi=2 Y

limi(x

primei)

|x2|2|x3|2|x4|2

int |x2||x1|


l1l2l3l4

=sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

|x2|2Y l3m3

(xprime3)

|x3|2Y l4m4

(xprime4)

|x4|2Cm1m2m3m4

l1l2l3l4

(minusl1 + l2 + l3 + l4 + 2)


1

)(39)





IS21 =sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

|x2|2Y l3m3

(xprime3)

|x3|2Y l4m4

(xprime4)

|x4|2Cm1m2m3m4

l1l2l3l4

(minusl1 + l2 + l3 + l4 + 2)


1

)IS22 =

sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

|x2|2Y l3m3

(xprime3)

|x3|2Y l4m4

(xprime4)

|x4|2Cm1m2m3m4

l1l2l3l4

(minusl1 + l2 + l3 + l4 + 2)


)(310)




primemicro3 x



I2 = IS21 + IR21 (311)


I3 =sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

int |x3||x2|


=sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4log

(|x3||x2|

)δ(l1 + l2 l3 + l4)

+sum

limil 6=0

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2

(|x3|l minus |x2|l

l


(312)

ndash 11 ndash

JHEP03(2016)165


IL32 =minussumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4log (|x2|) δ(l1 + l2 l3 + l4)

IS32 =sum

limil 6=0

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

(minus|x2|l

l


IL33 =sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4log (|x3|) δ(l1 + l2 l3 + l4)

IS33 =sum

limil 6=0

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

(|x3|l

l


(313)




I3 = IL32 + IS32 + IL33 + IS33 (314)



we know that

|x2| = max(r1 r2) =1

2(r1 + r2 + |r1 minus r2|)

|x3| = min(r3 r4) =1

2(r3 + r4 minus |r3 minus r4|) (315)



)(316)





I4 =sumlimi

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)


int |x4||x3|


=sum

limille0

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)

|x4|2


l minus 2



I43 =sum

limille0

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)

|x4|2

(minus|x3|lminus2

l minus 2


I44 =sum

limille0

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)

|x4|2

(|x4|lminus2

l minus 2

)(318)

ndash 12 ndash

JHEP03(2016)165




I4 = I43 + I44 (319)


I5 =sumlimi

Cm1m2m3m4

l1l2l3l4

prodi

Y limi(xi)

int infin|x4|


=sumlimil

Cm1m2m3m4

l1l2l3l4

prodi

Y limi(xi)

(|x4|minus4minusl

4 + l

)δ(l l1 + l2 + l3 + l4) (320)






C2 = ∆(∆minus 4) + l(l + 2) = minus3 (321)










and Libine [29]


result is

log(r3r2

)sumlimi

(Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

)δ(l1 + l2 l3 + l4) (41)


Φ(s t) = F0(s t) + log(s)F1(s t) (42)



ndash 13 ndash

JHEP03(2016)165





)spart

parts

(spartH

parts

)minus(


(tpartH

partt

)+ 4s

partH

parts(44)


C2 sΦ = 2 sΦ C2 sF1 = 2 sF1 (45)



for the integral is

1

2x213x224

F1(s t) (46)



plies that

2sumlimi

(Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

)δ(l1 + l2 l3 + l4) =

1

x213x224

F1(s t)

(47)




V+ otimes V+ = A[200] +

infinoplusk=1

D[k+2 k2 k2] (48)








P++++ defined by




ndash 14 ndash

JHEP03(2016)165



















1

2x213x224



prime24 P(x1 x2 x










ndash 15 ndash

JHEP03(2016)165





m1(x)Y l2

m2(x)Y l5

m5(x) (416)


(x1)Yl2m2


Y l1m1

(x1)otimes Y l2m2


(x)Y l2m2

(x)

Y l3m3

(xprime3)

|x3|2otimesY l4m4

(xprime4)

|x4|2rarr

Y l3m3

(xprime)

|x|2otimesY l4m4

(xprime)

|x|2(417)















m(K)Y ml (xprime)

sumlimi

(minusi)l12l1 l1

Y l1m1


Y l2m2

(x2)(i)l3

2l3 l3Y l3m3

(xprime3)(i)l4

2l4 l4Y l4m4

(xprime4)

P++minusminus

(Y m1l1




(K)vminus)

=sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

P++minusminus

(Y m1l1

(P )

2l1 l1v+ otimes

Y m2l2

(P )

2l2 l2v+ otimes

Y m3l3

(K)

2l3 l3vminus otimes

Y m4l4

(K)

2l4 l4vminus)

(418)








ndash 16 ndash

JHEP03(2016)165






F1 =1

λlog

((1 + ρs)


)(419)



sk

u2k+1[Qk(u) log(1 + u)]+ (420)



Qk(u) =

ksuml=0

bl(k)ul (421)

where

bl(k) =

(k

k minus l

)(2k + l

k minus l

)(422)








equivariant maps


I = I1 + I2 + I3 + I4 + I5



I(1) = (IS11 + IS21)

I(2) = (IS22 + IS32 + IL32)

I(3) = (IL33 + IS33 + IS43)

I(4) = (IS45 + IS55) (52)

ndash 17 ndash

JHEP03(2016)165




I = I(1) + I(2) + I(3) + I(4) (53)



jI(i) = minus2δij

prodk 6=j

1

(xk minus xi)2(54)









I(1) =sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

r22

Y l3m3

(xprime3)

r23

Y l4m4

(xprime4)



(l1 + l2 + l3 + l4 + 4)(minusl1 + l2 + l3 + l4 + 2)(55)




part2


(r21)A = 4A(A+ 1)(r21)Aminus1

part


part

partxmicro1Y l1m1

(x1)part


21)Aminus1Y l1

m1(x1) (56)


part2


I(1) =sumlimi


r22r23r

24

Y l1m1

(x1)Yl2m2

(xprime2)Yl3m3

(xprime3)Yl4m4


(57)


1


1

r22r23r

24

sumlimi

Y l2m2

(x1)Yl3m3

(x1)Yl4m4

(x1)Yl2m2

(xprime2)Yl3m3

(xprime3)Yl4m4

(xprime4)

=(l1 + 1)rl2+l3+l41

r22r23r

24

sumlimi

Cm1m2m3m4l1l2l3l4

Y l1m1

(x)Y l2m2

(xprime2)Yl3m3

(xprime3)Yl4m4

(xprime4) (58)

ndash 18 ndash

JHEP03(2016)165














I(2) =sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


1

l2+l3+l4minusl1+2

δ(l1 + l2 l3 + l4)

r23r24

+sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


1

r23r24

times



]


part

partxmicro2Y l2m2

(x2)part


22)Aminus1Y l2

m2(x2)

part

partxmicro2logr2 =

xmicro2r22

part

partxmicro2

part

partxmicro2logr2 =

2

r22(59)

We find

part2




r23r24

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


(510)


minus 2







ndash 19 ndash

JHEP03(2016)165


equivariant map


such that



x22x23x

24








23x

24


prime2)f(x1 x

prime3)f(x1 x

prime4)

(515)



1


(516)







part2



prime2)f(x1 x

prime3)f(x1 x

prime4) (518)


24I

(1)(x1 x2 x3 x4)


x22x23x

24 I(1)(x1 x2 x3 x4)rarr F1(e


While x23x24I








ndash 20 ndash

JHEP03(2016)165

implies that (part

partxmicro1+ x22

part

partxmicro2

1

x22+ x23

part

partxmicro3

1

x23+ x24

part

partxmicro4

1

x24

)F1 = 0 (521)


24I



24(I

(1)+I(2)+I(3)+I(4)) = x22x23x



partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)I = 0 (522)





23x

24I(x1 x2 x3 x4) =

x22x23x

24

x213x224

Φ(s t)

(523)



with


minus)=xprime21 x

prime22 xprime23


The x21x22x

23I













〈φ(x1)φ3(x2)φ(x3)φ(x4)〉 (528)




ndash 21 ndash

JHEP03(2016)165



twisted co-product










23x

24



















prime22 xprime23






ndash 22 ndash

JHEP03(2016)165





I2 =

intd4x5d

4x6xminus256

x215x225x

245x

226x

246x

236

(61)






















ndash 23 ndash

JHEP03(2016)165
































ndash 24 ndash

JHEP03(2016)165































Acknowledgments








this work was done

ndash 25 ndash

JHEP03(2016)165








1

2π2

intdS3Y l1

m1(x)Y l2

m2(x) = gm1m2

δl1l2

(l1 + 1)(A1)




gm1m2 = δm1m2


m2(x) = Y l

m1(x)

Y ml (x) = Y l

m(x) (A2)






(mLmR)

gm1m2 = δm1minusm2

Y ml (x) = Y l

minusm(x) (A3)



ables are

z = x1 + ix2 harr(

1

2

1

2


(minus1

2minus1

2

)w = x3 + ix4 harr

(1

2minus1

2


(minus1

21

2

)(A4)

We find that

Y l(l2l2) = (x1 + ix2)

l = zl (A5)

has normalization

1

2π2

intdS3(Y l

( l2 l2))lowast Y l

( l2 l2)

=1

(l + 1)(A6)


ndash 26 ndash

JHEP03(2016)165




(minusi)l

2llY lm(x)Y m

l (P )v+ (A7)


η(Y l1m1

(P )v+ Y m2l2

(K)vminus)

= δl1l2δm1m222l1(l1)2 (A8)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)= δl1l2δm1m2 (A9)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)=

1

2π2

intS3

Y m2l2

(xprime)


m1(x)) (A10)





Y lm(xprime)

xprime2


m(x) (A12)


η = f η


1


summ

1

y2Y lm(x)Y m


=infinsuml=0

summ

1

x2Y lm(xprime)Y m

l (y) if |x| gt |y| (A13)






ndash 27 ndash

JHEP03(2016)165


C2 = minus1

2LMNL


2MpqMqp (A14)






(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)eminusiP middotxv+

=

(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)ipart





Mpq = xppart

partxqminus xq

part

partxp(A16)


1

2MpqMqp(r

AY lm(x)) =

rA

2MpqMqpY


m(x) (A17)





m(x) (A19)

Thus we have

C2 rAY l




m(x) (A21)





F1 =1

λlog

((1 + ρs)


)(B1)

ndash 28 ndash

JHEP03(2016)165


of s to obtain

F1 =log(1+u)

u+

s


s2

u5(3u(2+u)minus 6 + 6u+ u2) log(1+u)

)+s3


)(B2)


F1 =

(minus1 +

u

2minus u2

3+u3


)+ s

(minus1

6+u

6minus 3u2


)+ s2

(minus1

30+

u

20minus 2u2


)+s3

(minus1

140+

u

70minus 5u2



The term at sk is

sk









in Pk(u) is

(2k)

(k)2(B6)


Qk(u) =ksuml=0

blul (B7)


ksuml=0

bl(minus1)l

(iminus l)= 0 (B8)




b1 =5b02 b2 =

20b09 b3 =

5b06 b4 =

5b042 b5 =

b0252

(B9)

ndash 29 ndash

JHEP03(2016)165



ksuml=0

MIlbl = 0 (B10)

where

MIl =(minus1)l




MIlbl (B12)



bl =sumI




bIb0 = 1 16bIb0 = 32 35 120bIb0 = 2 97 27 170

bIb0 =

5

2

20

95

6

5

42

1

252

bIb0 =

3

75

22

20

11

5

11

1

22

1

924

bIb0 =

7

2

63

13175

52

175

143

63

286

7

429

1

3432

(B14)



b1(k)b0 = k2

bk(k)b0 =k2

(2k)

bkminus1(k)b0 =k2

(2k)k(k + 1)

bkminus2(k)b0 =k2

(2k)

(k minus 1)k(k + 1)(k + 2)

4

bkminus3(k)b0 =k2

(2k)

(k

3

)(k + 3

3

)bkminusm(k)b0 =

k2

(2k)

(k

m

)(k +m

m

)(B15)

ndash 30 ndash

JHEP03(2016)165


so that

we have

b1(k) = k2(2k)

k2

bk(k) = 1



4

bkminus3(k) =

(k

3

)(k + 3

3

)bkminusm(k) =

(k

m

)(k +m

m

)(B16)


Pk(u) =

ksumi=1

aiuiminus1 (B17)



bl(minus1)iminusl

(iminus l)= 0 (B18)




k = 2 ~a = a1 ak = 6 3

k = 3 ~a =1

3times 60 60 11

k = 4 ~a =5

6times 84 126 52 5

k = 5 ~a =1

30times 7560 15120 9870 2310 137

k = 6 ~a =7

10times 13720 3300 2960 1140 174 7 (B19)

which leads to

a1(k) =(2k)

(k)2

ak(k) = 2h(k) = 2

ksuml=1

1

l(B20)




ndash 31 ndash

JHEP03(2016)165


grals






infinsumli=0

summi

infinsumt=0

sump


(x1)Yl2minusm2

(x2)Yl3m3

(xprime3)Yl4m4



t+ 1

=x23x

24

x213x224

1

λlog

((1 + ρs)


)(B21)



C l1 l2 lm1m2 m1+m2

=

langl12

l22

mL12

mL22

∣∣∣∣∣ l2

mL1 +mL2

2

ranglangl12

l22

mR12

mR22

∣∣∣∣∣ l2

mR1 +mR22






Yn12

n12n12

(x1)Yn22

n22n22



(x4) = zn11 zn2

2 zn33 zn4

4 (B22)


zn11 zn2

2 zn33 zn4


note that





that zn11 zn2

2 zn33 zn4


Yn12

n12n1minus2

2

(x1)Yn22

n2minus22

n22

(x2)Yn32


2

(x3)Yn42

minusn4minus22

minusn42

(x4)

=radicn1n2n3n4z

n1minus11 w1z

n2minus12 w2z

n3minus13 w3z

n4minus14 w4 (B24)


coefficients

Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B25)

Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B26)

ndash 32 ndash

JHEP03(2016)165


Cn12n22n1+n2

2n12n2minus2

2n1+n2minus2

2

=

radicn2

n1 + n2(B27)

Cn12n22n1+n2minus2

2n12n2minus2

2n1+n2minus2

2

= minusradic

n1n1 + n2

(B28)



n2minus12 w2z

n3minus13 w3z

n4minus14 w4 is

radicn1n2n3n4

[Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2 + 1

+Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2

]=


(B29)




section 5






parth








prime4vminus) (C1)



(2ixprime3microx

prime3 middot

part


part


)eiKmiddotx

prime3vminus

= ix23part

partxmicro3

1

x23eiKmiddotx

prime3vminus (C2)


parth


part

partxmicro2

(1

x22h

)minus x23

part

partxmicro3

(1

x23h

)minus x24

part

partxmicro4

(1

x24h

)(C3)

ndash 33 ndash

JHEP03(2016)165


x22x23x

24

(part

partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)[h

x22x23x

24

]= 0 (C4)


x1micropart

partx1νminus x1ν

part

partx1micro

)h= F1(Mmicroνe









x22x23x

24

(x1micro

part

partx1νminus x1ν

part


part

partx2νminus x2ν

part


part

partx3νminus x3ν

part

partx3micro

+x4micropart

partx4νminus x4ν

part

partx4micro

)[h

x22x23x

24

]= 0 (C6)


consider(x1 middot

part

partx1+ 3

)h = F1(e






x22x23x

24

(x1 middot

part

partx1+ x2 middot

part

partx2+ x3 middot

part

partx3+ x4 middot

part

partx4+ 12

)[h

x22x23x

24

]= 0 (C8)


h

x22x23x

24

=A




we have




ndash 34 ndash

JHEP03(2016)165





infinsumnm=0

im(minus1)n




where







minus (C14)

we find


and

infinsumn=0

(minus1)n




1


(C16)



=infinsum

n1n3n4=0

1



slot we get

α(P middot x2) =1


ndash 35 ndash

JHEP03(2016)165




=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)


=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)

(n1 + n3 + n4)



=infinsumn1

(minus1)n1




timesinfinsumn3

(minus1)n3




timesinfinsumn4

(minus1)n4





=1


1


1


=x22x

33x

24






References














ndash 36 ndash

JHEP03(2016)165










(2003) 013 [hep-th0212208] [INSPIRE]




(2014) 302 [arXiv14036646] [INSPIRE]






















(2015) 006 [arXiv150208005] [INSPIRE]






ndash 37 ndash

JHEP03(2016)165



[INSPIRE]

















(2015) 29FT01 [arXiv150500963] [INSPIRE]








[INSPIRE]














ndash 38 ndash

JHEP03(2016)165















ndash 39 ndash

Introduction























JHEP03(2016)165


I(x1 x2 x3 x4) =1

2x213x224

Φ(s t) (26)

where

Φ(s t) =1

λ


(t

s

)ln

(1 + ρt

1 + ρs

)+π2

3

)(27)

and

ρ =2



s =x212x

234

x213x224

t =x214x

223

x213x224

(28)



λ sim u+ s

(minus1minus 2

u

)+ s2

(minus 2

u3minus 2

u2

)sim u+ s

(minus1minus 2

u

)ρ sim s2

(2

u5+

1

u4

)+

s


1


s(29)



6minus 1

2log2(minusz) (210)


Φ(s t) =1

λ


(st

)log


(1 + ρs)

)(211)


notation


F0(s u) =1

λ



(1 + ρs)

)F1(s u) = minus 1

λlog


(1 + ρs)









=f(x1 x

prime4)f(x2 x

prime3)


ndash 7 ndash

JHEP03(2016)165

where








in section 4









J3

∣∣∣minus 1

2

rang= minus1

2

∣∣∣minus 1

2

rangJminus

∣∣∣minus 1

2

rang= 0 (214)





JminusJ2+

∣∣∣minus1

2

rang= 0 (215)




(2)

minus 12

The quotient

space Vminus 12V

(2)

minus 12


sequence

0rarr V(2)

minus 12


2V

(2)

minus 12

rarr 0 (216)









ndash 8 ndash

JHEP03(2016)165










(p2)+ rarr 0 (220)
























in section 5


I1 =sumllimi

prod4i=1 Y

limi(x

primei)

|x1|2|x2|2|x3|2|x4|2

int |x1|0


4prodi=1

Y mili

(x) (31)

ndash 9 ndash

JHEP03(2016)165

where d3x = 12π2dS



write it as

Cm1m2m3m4

l1l2l3l4=

suml5m5m6

Cm1m2l5l1l2m5

1

(l5 + 1)Cm3m4l5l3l4m6


(32)

where

Y m1l1

Y m2l2

=suml5m5

Cm1m2l5l1l2m5

Y m5l5

(33)




or equivalently












I1 =sumlimi

prodi Y

limi(x

primei)

|x1|2|x2|2|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

|x1|4+l1+l2+l3+l44 + l1 + l2 + l3 + l4

(38)




ndash 10 ndash

JHEP03(2016)165


region is

I2 =sumlimi

Y l1m1

(x1)prodi=2 Y

limi(x

primei)

|x2|2|x3|2|x4|2

int |x2||x1|


l1(xprime)Y m2

l2(x)Y m3

l3(x)Y m4

l4(x)

=sumlimi

Y l1m1

(x1)prodi=2 Y

limi(x

primei)

|x2|2|x3|2|x4|2

int |x2||x1|


l1l2l3l4

=sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

|x2|2Y l3m3

(xprime3)

|x3|2Y l4m4

(xprime4)

|x4|2Cm1m2m3m4

l1l2l3l4

(minusl1 + l2 + l3 + l4 + 2)


1

)(39)





IS21 =sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

|x2|2Y l3m3

(xprime3)

|x3|2Y l4m4

(xprime4)

|x4|2Cm1m2m3m4

l1l2l3l4

(minusl1 + l2 + l3 + l4 + 2)


1

)IS22 =

sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

|x2|2Y l3m3

(xprime3)

|x3|2Y l4m4

(xprime4)

|x4|2Cm1m2m3m4

l1l2l3l4

(minusl1 + l2 + l3 + l4 + 2)


)(310)




primemicro3 x



I2 = IS21 + IR21 (311)


I3 =sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

int |x3||x2|


=sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4log

(|x3||x2|

)δ(l1 + l2 l3 + l4)

+sum

limil 6=0

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2

(|x3|l minus |x2|l

l


(312)

ndash 11 ndash

JHEP03(2016)165


IL32 =minussumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4log (|x2|) δ(l1 + l2 l3 + l4)

IS32 =sum

limil 6=0

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

(minus|x2|l

l


IL33 =sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4log (|x3|) δ(l1 + l2 l3 + l4)

IS33 =sum

limil 6=0

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

(|x3|l

l


(313)




I3 = IL32 + IS32 + IL33 + IS33 (314)



we know that

|x2| = max(r1 r2) =1

2(r1 + r2 + |r1 minus r2|)

|x3| = min(r3 r4) =1

2(r3 + r4 minus |r3 minus r4|) (315)



)(316)





I4 =sumlimi

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)


int |x4||x3|


=sum

limille0

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)

|x4|2


l minus 2



I43 =sum

limille0

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)

|x4|2

(minus|x3|lminus2

l minus 2


I44 =sum

limille0

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)

|x4|2

(|x4|lminus2

l minus 2

)(318)

ndash 12 ndash

JHEP03(2016)165




I4 = I43 + I44 (319)


I5 =sumlimi

Cm1m2m3m4

l1l2l3l4

prodi

Y limi(xi)

int infin|x4|


=sumlimil

Cm1m2m3m4

l1l2l3l4

prodi

Y limi(xi)

(|x4|minus4minusl

4 + l

)δ(l l1 + l2 + l3 + l4) (320)






C2 = ∆(∆minus 4) + l(l + 2) = minus3 (321)










and Libine [29]


result is

log(r3r2

)sumlimi

(Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

)δ(l1 + l2 l3 + l4) (41)


Φ(s t) = F0(s t) + log(s)F1(s t) (42)



ndash 13 ndash

JHEP03(2016)165





)spart

parts

(spartH

parts

)minus(


(tpartH

partt

)+ 4s

partH

parts(44)


C2 sΦ = 2 sΦ C2 sF1 = 2 sF1 (45)



for the integral is

1

2x213x224

F1(s t) (46)



plies that

2sumlimi

(Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

)δ(l1 + l2 l3 + l4) =

1

x213x224

F1(s t)

(47)




V+ otimes V+ = A[200] +

infinoplusk=1

D[k+2 k2 k2] (48)








P++++ defined by




ndash 14 ndash

JHEP03(2016)165



















1

2x213x224



prime24 P(x1 x2 x










ndash 15 ndash

JHEP03(2016)165





m1(x)Y l2

m2(x)Y l5

m5(x) (416)


(x1)Yl2m2


Y l1m1

(x1)otimes Y l2m2


(x)Y l2m2

(x)

Y l3m3

(xprime3)

|x3|2otimesY l4m4

(xprime4)

|x4|2rarr

Y l3m3

(xprime)

|x|2otimesY l4m4

(xprime)

|x|2(417)















m(K)Y ml (xprime)

sumlimi

(minusi)l12l1 l1

Y l1m1


Y l2m2

(x2)(i)l3

2l3 l3Y l3m3

(xprime3)(i)l4

2l4 l4Y l4m4

(xprime4)

P++minusminus

(Y m1l1




(K)vminus)

=sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

P++minusminus

(Y m1l1

(P )

2l1 l1v+ otimes

Y m2l2

(P )

2l2 l2v+ otimes

Y m3l3

(K)

2l3 l3vminus otimes

Y m4l4

(K)

2l4 l4vminus)

(418)








ndash 16 ndash

JHEP03(2016)165






F1 =1

λlog

((1 + ρs)


)(419)



sk

u2k+1[Qk(u) log(1 + u)]+ (420)



Qk(u) =

ksuml=0

bl(k)ul (421)

where

bl(k) =

(k

k minus l

)(2k + l

k minus l

)(422)








equivariant maps


I = I1 + I2 + I3 + I4 + I5



I(1) = (IS11 + IS21)

I(2) = (IS22 + IS32 + IL32)

I(3) = (IL33 + IS33 + IS43)

I(4) = (IS45 + IS55) (52)

ndash 17 ndash

JHEP03(2016)165




I = I(1) + I(2) + I(3) + I(4) (53)



jI(i) = minus2δij

prodk 6=j

1

(xk minus xi)2(54)









I(1) =sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

r22

Y l3m3

(xprime3)

r23

Y l4m4

(xprime4)



(l1 + l2 + l3 + l4 + 4)(minusl1 + l2 + l3 + l4 + 2)(55)




part2


(r21)A = 4A(A+ 1)(r21)Aminus1

part


part

partxmicro1Y l1m1

(x1)part


21)Aminus1Y l1

m1(x1) (56)


part2


I(1) =sumlimi


r22r23r

24

Y l1m1

(x1)Yl2m2

(xprime2)Yl3m3

(xprime3)Yl4m4


(57)


1


1

r22r23r

24

sumlimi

Y l2m2

(x1)Yl3m3

(x1)Yl4m4

(x1)Yl2m2

(xprime2)Yl3m3

(xprime3)Yl4m4

(xprime4)

=(l1 + 1)rl2+l3+l41

r22r23r

24

sumlimi

Cm1m2m3m4l1l2l3l4

Y l1m1

(x)Y l2m2

(xprime2)Yl3m3

(xprime3)Yl4m4

(xprime4) (58)

ndash 18 ndash

JHEP03(2016)165














I(2) =sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


1

l2+l3+l4minusl1+2

δ(l1 + l2 l3 + l4)

r23r24

+sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


1

r23r24

times



]


part

partxmicro2Y l2m2

(x2)part


22)Aminus1Y l2

m2(x2)

part

partxmicro2logr2 =

xmicro2r22

part

partxmicro2

part

partxmicro2logr2 =

2

r22(59)

We find

part2




r23r24

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


(510)


minus 2







ndash 19 ndash

JHEP03(2016)165


equivariant map


such that



x22x23x

24








23x

24


prime2)f(x1 x

prime3)f(x1 x

prime4)

(515)



1


(516)







part2



prime2)f(x1 x

prime3)f(x1 x

prime4) (518)


24I

(1)(x1 x2 x3 x4)


x22x23x

24 I(1)(x1 x2 x3 x4)rarr F1(e


While x23x24I








ndash 20 ndash

JHEP03(2016)165

implies that (part

partxmicro1+ x22

part

partxmicro2

1

x22+ x23

part

partxmicro3

1

x23+ x24

part

partxmicro4

1

x24

)F1 = 0 (521)


24I



24(I

(1)+I(2)+I(3)+I(4)) = x22x23x



partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)I = 0 (522)





23x

24I(x1 x2 x3 x4) =

x22x23x

24

x213x224

Φ(s t)

(523)



with


minus)=xprime21 x

prime22 xprime23


The x21x22x

23I













〈φ(x1)φ3(x2)φ(x3)φ(x4)〉 (528)




ndash 21 ndash

JHEP03(2016)165



twisted co-product










23x

24



















prime22 xprime23






ndash 22 ndash

JHEP03(2016)165





I2 =

intd4x5d

4x6xminus256

x215x225x

245x

226x

246x

236

(61)






















ndash 23 ndash

JHEP03(2016)165
































ndash 24 ndash

JHEP03(2016)165































Acknowledgments








this work was done

ndash 25 ndash

JHEP03(2016)165








1

2π2

intdS3Y l1

m1(x)Y l2

m2(x) = gm1m2

δl1l2

(l1 + 1)(A1)




gm1m2 = δm1m2


m2(x) = Y l

m1(x)

Y ml (x) = Y l

m(x) (A2)






(mLmR)

gm1m2 = δm1minusm2

Y ml (x) = Y l

minusm(x) (A3)



ables are

z = x1 + ix2 harr(

1

2

1

2


(minus1

2minus1

2

)w = x3 + ix4 harr

(1

2minus1

2


(minus1

21

2

)(A4)

We find that

Y l(l2l2) = (x1 + ix2)

l = zl (A5)

has normalization

1

2π2

intdS3(Y l

( l2 l2))lowast Y l

( l2 l2)

=1

(l + 1)(A6)


ndash 26 ndash

JHEP03(2016)165




(minusi)l

2llY lm(x)Y m

l (P )v+ (A7)


η(Y l1m1

(P )v+ Y m2l2

(K)vminus)

= δl1l2δm1m222l1(l1)2 (A8)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)= δl1l2δm1m2 (A9)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)=

1

2π2

intS3

Y m2l2

(xprime)


m1(x)) (A10)





Y lm(xprime)

xprime2


m(x) (A12)


η = f η


1


summ

1

y2Y lm(x)Y m


=infinsuml=0

summ

1

x2Y lm(xprime)Y m

l (y) if |x| gt |y| (A13)






ndash 27 ndash

JHEP03(2016)165


C2 = minus1

2LMNL


2MpqMqp (A14)






(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)eminusiP middotxv+

=

(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)ipart





Mpq = xppart

partxqminus xq

part

partxp(A16)


1

2MpqMqp(r

AY lm(x)) =

rA

2MpqMqpY


m(x) (A17)





m(x) (A19)

Thus we have

C2 rAY l




m(x) (A21)





F1 =1

λlog

((1 + ρs)


)(B1)

ndash 28 ndash

JHEP03(2016)165


of s to obtain

F1 =log(1+u)

u+

s


s2

u5(3u(2+u)minus 6 + 6u+ u2) log(1+u)

)+s3


)(B2)


F1 =

(minus1 +

u

2minus u2

3+u3


)+ s

(minus1

6+u

6minus 3u2


)+ s2

(minus1

30+

u

20minus 2u2


)+s3

(minus1

140+

u

70minus 5u2



The term at sk is

sk









in Pk(u) is

(2k)

(k)2(B6)


Qk(u) =ksuml=0

blul (B7)


ksuml=0

bl(minus1)l

(iminus l)= 0 (B8)




b1 =5b02 b2 =

20b09 b3 =

5b06 b4 =

5b042 b5 =

b0252

(B9)

ndash 29 ndash

JHEP03(2016)165



ksuml=0

MIlbl = 0 (B10)

where

MIl =(minus1)l




MIlbl (B12)



bl =sumI




bIb0 = 1 16bIb0 = 32 35 120bIb0 = 2 97 27 170

bIb0 =

5

2

20

95

6

5

42

1

252

bIb0 =

3

75

22

20

11

5

11

1

22

1

924

bIb0 =

7

2

63

13175

52

175

143

63

286

7

429

1

3432

(B14)



b1(k)b0 = k2

bk(k)b0 =k2

(2k)

bkminus1(k)b0 =k2

(2k)k(k + 1)

bkminus2(k)b0 =k2

(2k)

(k minus 1)k(k + 1)(k + 2)

4

bkminus3(k)b0 =k2

(2k)

(k

3

)(k + 3

3

)bkminusm(k)b0 =

k2

(2k)

(k

m

)(k +m

m

)(B15)

ndash 30 ndash

JHEP03(2016)165


so that

we have

b1(k) = k2(2k)

k2

bk(k) = 1



4

bkminus3(k) =

(k

3

)(k + 3

3

)bkminusm(k) =

(k

m

)(k +m

m

)(B16)


Pk(u) =

ksumi=1

aiuiminus1 (B17)



bl(minus1)iminusl

(iminus l)= 0 (B18)




k = 2 ~a = a1 ak = 6 3

k = 3 ~a =1

3times 60 60 11

k = 4 ~a =5

6times 84 126 52 5

k = 5 ~a =1

30times 7560 15120 9870 2310 137

k = 6 ~a =7

10times 13720 3300 2960 1140 174 7 (B19)

which leads to

a1(k) =(2k)

(k)2

ak(k) = 2h(k) = 2

ksuml=1

1

l(B20)




ndash 31 ndash

JHEP03(2016)165


grals






infinsumli=0

summi

infinsumt=0

sump


(x1)Yl2minusm2

(x2)Yl3m3

(xprime3)Yl4m4



t+ 1

=x23x

24

x213x224

1

λlog

((1 + ρs)


)(B21)



C l1 l2 lm1m2 m1+m2

=

langl12

l22

mL12

mL22

∣∣∣∣∣ l2

mL1 +mL2

2

ranglangl12

l22

mR12

mR22

∣∣∣∣∣ l2

mR1 +mR22






Yn12

n12n12

(x1)Yn22

n22n22



(x4) = zn11 zn2

2 zn33 zn4

4 (B22)


zn11 zn2

2 zn33 zn4


note that





that zn11 zn2

2 zn33 zn4


Yn12

n12n1minus2

2

(x1)Yn22

n2minus22

n22

(x2)Yn32


2

(x3)Yn42

minusn4minus22

minusn42

(x4)

=radicn1n2n3n4z

n1minus11 w1z

n2minus12 w2z

n3minus13 w3z

n4minus14 w4 (B24)


coefficients

Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B25)

Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B26)

ndash 32 ndash

JHEP03(2016)165


Cn12n22n1+n2

2n12n2minus2

2n1+n2minus2

2

=

radicn2

n1 + n2(B27)

Cn12n22n1+n2minus2

2n12n2minus2

2n1+n2minus2

2

= minusradic

n1n1 + n2

(B28)



n2minus12 w2z

n3minus13 w3z

n4minus14 w4 is

radicn1n2n3n4

[Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2 + 1

+Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2

]=


(B29)




section 5






parth








prime4vminus) (C1)



(2ixprime3microx

prime3 middot

part


part


)eiKmiddotx

prime3vminus

= ix23part

partxmicro3

1

x23eiKmiddotx

prime3vminus (C2)


parth


part

partxmicro2

(1

x22h

)minus x23

part

partxmicro3

(1

x23h

)minus x24

part

partxmicro4

(1

x24h

)(C3)

ndash 33 ndash

JHEP03(2016)165


x22x23x

24

(part

partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)[h

x22x23x

24

]= 0 (C4)


x1micropart

partx1νminus x1ν

part

partx1micro

)h= F1(Mmicroνe









x22x23x

24

(x1micro

part

partx1νminus x1ν

part


part

partx2νminus x2ν

part


part

partx3νminus x3ν

part

partx3micro

+x4micropart

partx4νminus x4ν

part

partx4micro

)[h

x22x23x

24

]= 0 (C6)


consider(x1 middot

part

partx1+ 3

)h = F1(e






x22x23x

24

(x1 middot

part

partx1+ x2 middot

part

partx2+ x3 middot

part

partx3+ x4 middot

part

partx4+ 12

)[h

x22x23x

24

]= 0 (C8)


h

x22x23x

24

=A




we have




ndash 34 ndash

JHEP03(2016)165





infinsumnm=0

im(minus1)n




where







minus (C14)

we find


and

infinsumn=0

(minus1)n




1


(C16)



=infinsum

n1n3n4=0

1



slot we get

α(P middot x2) =1


ndash 35 ndash

JHEP03(2016)165




=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)


=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)

(n1 + n3 + n4)



=infinsumn1

(minus1)n1




timesinfinsumn3

(minus1)n3




timesinfinsumn4

(minus1)n4





=1


1


1


=x22x

33x

24






References














ndash 36 ndash

JHEP03(2016)165










(2003) 013 [hep-th0212208] [INSPIRE]




(2014) 302 [arXiv14036646] [INSPIRE]






















(2015) 006 [arXiv150208005] [INSPIRE]






ndash 37 ndash

JHEP03(2016)165



[INSPIRE]

















(2015) 29FT01 [arXiv150500963] [INSPIRE]








[INSPIRE]














ndash 38 ndash

JHEP03(2016)165















ndash 39 ndash

Introduction























JHEP03(2016)165

where








in section 4









J3

∣∣∣minus 1

2

rang= minus1

2

∣∣∣minus 1

2

rangJminus

∣∣∣minus 1

2

rang= 0 (214)





JminusJ2+

∣∣∣minus1

2

rang= 0 (215)




(2)

minus 12

The quotient

space Vminus 12V

(2)

minus 12


sequence

0rarr V(2)

minus 12


2V

(2)

minus 12

rarr 0 (216)









ndash 8 ndash

JHEP03(2016)165










(p2)+ rarr 0 (220)
























in section 5


I1 =sumllimi

prod4i=1 Y

limi(x

primei)

|x1|2|x2|2|x3|2|x4|2

int |x1|0


4prodi=1

Y mili

(x) (31)

ndash 9 ndash

JHEP03(2016)165

where d3x = 12π2dS



write it as

Cm1m2m3m4

l1l2l3l4=

suml5m5m6

Cm1m2l5l1l2m5

1

(l5 + 1)Cm3m4l5l3l4m6


(32)

where

Y m1l1

Y m2l2

=suml5m5

Cm1m2l5l1l2m5

Y m5l5

(33)




or equivalently












I1 =sumlimi

prodi Y

limi(x

primei)

|x1|2|x2|2|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

|x1|4+l1+l2+l3+l44 + l1 + l2 + l3 + l4

(38)




ndash 10 ndash

JHEP03(2016)165


region is

I2 =sumlimi

Y l1m1

(x1)prodi=2 Y

limi(x

primei)

|x2|2|x3|2|x4|2

int |x2||x1|


l1(xprime)Y m2

l2(x)Y m3

l3(x)Y m4

l4(x)

=sumlimi

Y l1m1

(x1)prodi=2 Y

limi(x

primei)

|x2|2|x3|2|x4|2

int |x2||x1|


l1l2l3l4

=sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

|x2|2Y l3m3

(xprime3)

|x3|2Y l4m4

(xprime4)

|x4|2Cm1m2m3m4

l1l2l3l4

(minusl1 + l2 + l3 + l4 + 2)


1

)(39)





IS21 =sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

|x2|2Y l3m3

(xprime3)

|x3|2Y l4m4

(xprime4)

|x4|2Cm1m2m3m4

l1l2l3l4

(minusl1 + l2 + l3 + l4 + 2)


1

)IS22 =

sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

|x2|2Y l3m3

(xprime3)

|x3|2Y l4m4

(xprime4)

|x4|2Cm1m2m3m4

l1l2l3l4

(minusl1 + l2 + l3 + l4 + 2)


)(310)




primemicro3 x



I2 = IS21 + IR21 (311)


I3 =sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

int |x3||x2|


=sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4log

(|x3||x2|

)δ(l1 + l2 l3 + l4)

+sum

limil 6=0

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2

(|x3|l minus |x2|l

l


(312)

ndash 11 ndash

JHEP03(2016)165


IL32 =minussumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4log (|x2|) δ(l1 + l2 l3 + l4)

IS32 =sum

limil 6=0

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

(minus|x2|l

l


IL33 =sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4log (|x3|) δ(l1 + l2 l3 + l4)

IS33 =sum

limil 6=0

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

(|x3|l

l


(313)




I3 = IL32 + IS32 + IL33 + IS33 (314)



we know that

|x2| = max(r1 r2) =1

2(r1 + r2 + |r1 minus r2|)

|x3| = min(r3 r4) =1

2(r3 + r4 minus |r3 minus r4|) (315)



)(316)





I4 =sumlimi

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)


int |x4||x3|


=sum

limille0

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)

|x4|2


l minus 2



I43 =sum

limille0

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)

|x4|2

(minus|x3|lminus2

l minus 2


I44 =sum

limille0

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)

|x4|2

(|x4|lminus2

l minus 2

)(318)

ndash 12 ndash

JHEP03(2016)165




I4 = I43 + I44 (319)


I5 =sumlimi

Cm1m2m3m4

l1l2l3l4

prodi

Y limi(xi)

int infin|x4|


=sumlimil

Cm1m2m3m4

l1l2l3l4

prodi

Y limi(xi)

(|x4|minus4minusl

4 + l

)δ(l l1 + l2 + l3 + l4) (320)






C2 = ∆(∆minus 4) + l(l + 2) = minus3 (321)










and Libine [29]


result is

log(r3r2

)sumlimi

(Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

)δ(l1 + l2 l3 + l4) (41)


Φ(s t) = F0(s t) + log(s)F1(s t) (42)



ndash 13 ndash

JHEP03(2016)165





)spart

parts

(spartH

parts

)minus(


(tpartH

partt

)+ 4s

partH

parts(44)


C2 sΦ = 2 sΦ C2 sF1 = 2 sF1 (45)



for the integral is

1

2x213x224

F1(s t) (46)



plies that

2sumlimi

(Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

)δ(l1 + l2 l3 + l4) =

1

x213x224

F1(s t)

(47)




V+ otimes V+ = A[200] +

infinoplusk=1

D[k+2 k2 k2] (48)








P++++ defined by




ndash 14 ndash

JHEP03(2016)165



















1

2x213x224



prime24 P(x1 x2 x










ndash 15 ndash

JHEP03(2016)165





m1(x)Y l2

m2(x)Y l5

m5(x) (416)


(x1)Yl2m2


Y l1m1

(x1)otimes Y l2m2


(x)Y l2m2

(x)

Y l3m3

(xprime3)

|x3|2otimesY l4m4

(xprime4)

|x4|2rarr

Y l3m3

(xprime)

|x|2otimesY l4m4

(xprime)

|x|2(417)















m(K)Y ml (xprime)

sumlimi

(minusi)l12l1 l1

Y l1m1


Y l2m2

(x2)(i)l3

2l3 l3Y l3m3

(xprime3)(i)l4

2l4 l4Y l4m4

(xprime4)

P++minusminus

(Y m1l1




(K)vminus)

=sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

P++minusminus

(Y m1l1

(P )

2l1 l1v+ otimes

Y m2l2

(P )

2l2 l2v+ otimes

Y m3l3

(K)

2l3 l3vminus otimes

Y m4l4

(K)

2l4 l4vminus)

(418)








ndash 16 ndash

JHEP03(2016)165






F1 =1

λlog

((1 + ρs)


)(419)



sk

u2k+1[Qk(u) log(1 + u)]+ (420)



Qk(u) =

ksuml=0

bl(k)ul (421)

where

bl(k) =

(k

k minus l

)(2k + l

k minus l

)(422)








equivariant maps


I = I1 + I2 + I3 + I4 + I5



I(1) = (IS11 + IS21)

I(2) = (IS22 + IS32 + IL32)

I(3) = (IL33 + IS33 + IS43)

I(4) = (IS45 + IS55) (52)

ndash 17 ndash

JHEP03(2016)165




I = I(1) + I(2) + I(3) + I(4) (53)



jI(i) = minus2δij

prodk 6=j

1

(xk minus xi)2(54)









I(1) =sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

r22

Y l3m3

(xprime3)

r23

Y l4m4

(xprime4)



(l1 + l2 + l3 + l4 + 4)(minusl1 + l2 + l3 + l4 + 2)(55)




part2


(r21)A = 4A(A+ 1)(r21)Aminus1

part


part

partxmicro1Y l1m1

(x1)part


21)Aminus1Y l1

m1(x1) (56)


part2


I(1) =sumlimi


r22r23r

24

Y l1m1

(x1)Yl2m2

(xprime2)Yl3m3

(xprime3)Yl4m4


(57)


1


1

r22r23r

24

sumlimi

Y l2m2

(x1)Yl3m3

(x1)Yl4m4

(x1)Yl2m2

(xprime2)Yl3m3

(xprime3)Yl4m4

(xprime4)

=(l1 + 1)rl2+l3+l41

r22r23r

24

sumlimi

Cm1m2m3m4l1l2l3l4

Y l1m1

(x)Y l2m2

(xprime2)Yl3m3

(xprime3)Yl4m4

(xprime4) (58)

ndash 18 ndash

JHEP03(2016)165














I(2) =sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


1

l2+l3+l4minusl1+2

δ(l1 + l2 l3 + l4)

r23r24

+sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


1

r23r24

times



]


part

partxmicro2Y l2m2

(x2)part


22)Aminus1Y l2

m2(x2)

part

partxmicro2logr2 =

xmicro2r22

part

partxmicro2

part

partxmicro2logr2 =

2

r22(59)

We find

part2




r23r24

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


(510)


minus 2







ndash 19 ndash

JHEP03(2016)165


equivariant map


such that



x22x23x

24








23x

24


prime2)f(x1 x

prime3)f(x1 x

prime4)

(515)



1


(516)







part2



prime2)f(x1 x

prime3)f(x1 x

prime4) (518)


24I

(1)(x1 x2 x3 x4)


x22x23x

24 I(1)(x1 x2 x3 x4)rarr F1(e


While x23x24I








ndash 20 ndash

JHEP03(2016)165

implies that (part

partxmicro1+ x22

part

partxmicro2

1

x22+ x23

part

partxmicro3

1

x23+ x24

part

partxmicro4

1

x24

)F1 = 0 (521)


24I



24(I

(1)+I(2)+I(3)+I(4)) = x22x23x



partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)I = 0 (522)





23x

24I(x1 x2 x3 x4) =

x22x23x

24

x213x224

Φ(s t)

(523)



with


minus)=xprime21 x

prime22 xprime23


The x21x22x

23I













〈φ(x1)φ3(x2)φ(x3)φ(x4)〉 (528)




ndash 21 ndash

JHEP03(2016)165



twisted co-product










23x

24



















prime22 xprime23






ndash 22 ndash

JHEP03(2016)165





I2 =

intd4x5d

4x6xminus256

x215x225x

245x

226x

246x

236

(61)






















ndash 23 ndash

JHEP03(2016)165
































ndash 24 ndash

JHEP03(2016)165































Acknowledgments








this work was done

ndash 25 ndash

JHEP03(2016)165








1

2π2

intdS3Y l1

m1(x)Y l2

m2(x) = gm1m2

δl1l2

(l1 + 1)(A1)




gm1m2 = δm1m2


m2(x) = Y l

m1(x)

Y ml (x) = Y l

m(x) (A2)






(mLmR)

gm1m2 = δm1minusm2

Y ml (x) = Y l

minusm(x) (A3)



ables are

z = x1 + ix2 harr(

1

2

1

2


(minus1

2minus1

2

)w = x3 + ix4 harr

(1

2minus1

2


(minus1

21

2

)(A4)

We find that

Y l(l2l2) = (x1 + ix2)

l = zl (A5)

has normalization

1

2π2

intdS3(Y l

( l2 l2))lowast Y l

( l2 l2)

=1

(l + 1)(A6)


ndash 26 ndash

JHEP03(2016)165




(minusi)l

2llY lm(x)Y m

l (P )v+ (A7)


η(Y l1m1

(P )v+ Y m2l2

(K)vminus)

= δl1l2δm1m222l1(l1)2 (A8)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)= δl1l2δm1m2 (A9)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)=

1

2π2

intS3

Y m2l2

(xprime)


m1(x)) (A10)





Y lm(xprime)

xprime2


m(x) (A12)


η = f η


1


summ

1

y2Y lm(x)Y m


=infinsuml=0

summ

1

x2Y lm(xprime)Y m

l (y) if |x| gt |y| (A13)






ndash 27 ndash

JHEP03(2016)165


C2 = minus1

2LMNL


2MpqMqp (A14)






(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)eminusiP middotxv+

=

(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)ipart





Mpq = xppart

partxqminus xq

part

partxp(A16)


1

2MpqMqp(r

AY lm(x)) =

rA

2MpqMqpY


m(x) (A17)





m(x) (A19)

Thus we have

C2 rAY l




m(x) (A21)





F1 =1

λlog

((1 + ρs)


)(B1)

ndash 28 ndash

JHEP03(2016)165


of s to obtain

F1 =log(1+u)

u+

s


s2

u5(3u(2+u)minus 6 + 6u+ u2) log(1+u)

)+s3


)(B2)


F1 =

(minus1 +

u

2minus u2

3+u3


)+ s

(minus1

6+u

6minus 3u2


)+ s2

(minus1

30+

u

20minus 2u2


)+s3

(minus1

140+

u

70minus 5u2



The term at sk is

sk









in Pk(u) is

(2k)

(k)2(B6)


Qk(u) =ksuml=0

blul (B7)


ksuml=0

bl(minus1)l

(iminus l)= 0 (B8)




b1 =5b02 b2 =

20b09 b3 =

5b06 b4 =

5b042 b5 =

b0252

(B9)

ndash 29 ndash

JHEP03(2016)165



ksuml=0

MIlbl = 0 (B10)

where

MIl =(minus1)l




MIlbl (B12)



bl =sumI




bIb0 = 1 16bIb0 = 32 35 120bIb0 = 2 97 27 170

bIb0 =

5

2

20

95

6

5

42

1

252

bIb0 =

3

75

22

20

11

5

11

1

22

1

924

bIb0 =

7

2

63

13175

52

175

143

63

286

7

429

1

3432

(B14)



b1(k)b0 = k2

bk(k)b0 =k2

(2k)

bkminus1(k)b0 =k2

(2k)k(k + 1)

bkminus2(k)b0 =k2

(2k)

(k minus 1)k(k + 1)(k + 2)

4

bkminus3(k)b0 =k2

(2k)

(k

3

)(k + 3

3

)bkminusm(k)b0 =

k2

(2k)

(k

m

)(k +m

m

)(B15)

ndash 30 ndash

JHEP03(2016)165


so that

we have

b1(k) = k2(2k)

k2

bk(k) = 1



4

bkminus3(k) =

(k

3

)(k + 3

3

)bkminusm(k) =

(k

m

)(k +m

m

)(B16)


Pk(u) =

ksumi=1

aiuiminus1 (B17)



bl(minus1)iminusl

(iminus l)= 0 (B18)




k = 2 ~a = a1 ak = 6 3

k = 3 ~a =1

3times 60 60 11

k = 4 ~a =5

6times 84 126 52 5

k = 5 ~a =1

30times 7560 15120 9870 2310 137

k = 6 ~a =7

10times 13720 3300 2960 1140 174 7 (B19)

which leads to

a1(k) =(2k)

(k)2

ak(k) = 2h(k) = 2

ksuml=1

1

l(B20)




ndash 31 ndash

JHEP03(2016)165


grals






infinsumli=0

summi

infinsumt=0

sump


(x1)Yl2minusm2

(x2)Yl3m3

(xprime3)Yl4m4



t+ 1

=x23x

24

x213x224

1

λlog

((1 + ρs)


)(B21)



C l1 l2 lm1m2 m1+m2

=

langl12

l22

mL12

mL22

∣∣∣∣∣ l2

mL1 +mL2

2

ranglangl12

l22

mR12

mR22

∣∣∣∣∣ l2

mR1 +mR22






Yn12

n12n12

(x1)Yn22

n22n22



(x4) = zn11 zn2

2 zn33 zn4

4 (B22)


zn11 zn2

2 zn33 zn4


note that





that zn11 zn2

2 zn33 zn4


Yn12

n12n1minus2

2

(x1)Yn22

n2minus22

n22

(x2)Yn32


2

(x3)Yn42

minusn4minus22

minusn42

(x4)

=radicn1n2n3n4z

n1minus11 w1z

n2minus12 w2z

n3minus13 w3z

n4minus14 w4 (B24)


coefficients

Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B25)

Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B26)

ndash 32 ndash

JHEP03(2016)165


Cn12n22n1+n2

2n12n2minus2

2n1+n2minus2

2

=

radicn2

n1 + n2(B27)

Cn12n22n1+n2minus2

2n12n2minus2

2n1+n2minus2

2

= minusradic

n1n1 + n2

(B28)



n2minus12 w2z

n3minus13 w3z

n4minus14 w4 is

radicn1n2n3n4

[Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2 + 1

+Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2

]=


(B29)




section 5






parth








prime4vminus) (C1)



(2ixprime3microx

prime3 middot

part


part


)eiKmiddotx

prime3vminus

= ix23part

partxmicro3

1

x23eiKmiddotx

prime3vminus (C2)


parth


part

partxmicro2

(1

x22h

)minus x23

part

partxmicro3

(1

x23h

)minus x24

part

partxmicro4

(1

x24h

)(C3)

ndash 33 ndash

JHEP03(2016)165


x22x23x

24

(part

partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)[h

x22x23x

24

]= 0 (C4)


x1micropart

partx1νminus x1ν

part

partx1micro

)h= F1(Mmicroνe









x22x23x

24

(x1micro

part

partx1νminus x1ν

part


part

partx2νminus x2ν

part


part

partx3νminus x3ν

part

partx3micro

+x4micropart

partx4νminus x4ν

part

partx4micro

)[h

x22x23x

24

]= 0 (C6)


consider(x1 middot

part

partx1+ 3

)h = F1(e






x22x23x

24

(x1 middot

part

partx1+ x2 middot

part

partx2+ x3 middot

part

partx3+ x4 middot

part

partx4+ 12

)[h

x22x23x

24

]= 0 (C8)


h

x22x23x

24

=A




we have




ndash 34 ndash

JHEP03(2016)165





infinsumnm=0

im(minus1)n




where







minus (C14)

we find


and

infinsumn=0

(minus1)n




1


(C16)



=infinsum

n1n3n4=0

1



slot we get

α(P middot x2) =1


ndash 35 ndash

JHEP03(2016)165




=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)


=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)

(n1 + n3 + n4)



=infinsumn1

(minus1)n1




timesinfinsumn3

(minus1)n3




timesinfinsumn4

(minus1)n4





=1


1


1


=x22x

33x

24






References














ndash 36 ndash

JHEP03(2016)165










(2003) 013 [hep-th0212208] [INSPIRE]




(2014) 302 [arXiv14036646] [INSPIRE]






















(2015) 006 [arXiv150208005] [INSPIRE]






ndash 37 ndash

JHEP03(2016)165



[INSPIRE]

















(2015) 29FT01 [arXiv150500963] [INSPIRE]








[INSPIRE]














ndash 38 ndash

JHEP03(2016)165















ndash 39 ndash

Introduction























JHEP03(2016)165










(p2)+ rarr 0 (220)
























in section 5


I1 =sumllimi

prod4i=1 Y

limi(x

primei)

|x1|2|x2|2|x3|2|x4|2

int |x1|0


4prodi=1

Y mili

(x) (31)

ndash 9 ndash

JHEP03(2016)165

where d3x = 12π2dS



write it as

Cm1m2m3m4

l1l2l3l4=

suml5m5m6

Cm1m2l5l1l2m5

1

(l5 + 1)Cm3m4l5l3l4m6


(32)

where

Y m1l1

Y m2l2

=suml5m5

Cm1m2l5l1l2m5

Y m5l5

(33)




or equivalently












I1 =sumlimi

prodi Y

limi(x

primei)

|x1|2|x2|2|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

|x1|4+l1+l2+l3+l44 + l1 + l2 + l3 + l4

(38)




ndash 10 ndash

JHEP03(2016)165


region is

I2 =sumlimi

Y l1m1

(x1)prodi=2 Y

limi(x

primei)

|x2|2|x3|2|x4|2

int |x2||x1|


l1(xprime)Y m2

l2(x)Y m3

l3(x)Y m4

l4(x)

=sumlimi

Y l1m1

(x1)prodi=2 Y

limi(x

primei)

|x2|2|x3|2|x4|2

int |x2||x1|


l1l2l3l4

=sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

|x2|2Y l3m3

(xprime3)

|x3|2Y l4m4

(xprime4)

|x4|2Cm1m2m3m4

l1l2l3l4

(minusl1 + l2 + l3 + l4 + 2)


1

)(39)





IS21 =sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

|x2|2Y l3m3

(xprime3)

|x3|2Y l4m4

(xprime4)

|x4|2Cm1m2m3m4

l1l2l3l4

(minusl1 + l2 + l3 + l4 + 2)


1

)IS22 =

sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

|x2|2Y l3m3

(xprime3)

|x3|2Y l4m4

(xprime4)

|x4|2Cm1m2m3m4

l1l2l3l4

(minusl1 + l2 + l3 + l4 + 2)


)(310)




primemicro3 x



I2 = IS21 + IR21 (311)


I3 =sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

int |x3||x2|


=sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4log

(|x3||x2|

)δ(l1 + l2 l3 + l4)

+sum

limil 6=0

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2

(|x3|l minus |x2|l

l


(312)

ndash 11 ndash

JHEP03(2016)165


IL32 =minussumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4log (|x2|) δ(l1 + l2 l3 + l4)

IS32 =sum

limil 6=0

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

(minus|x2|l

l


IL33 =sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4log (|x3|) δ(l1 + l2 l3 + l4)

IS33 =sum

limil 6=0

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

(|x3|l

l


(313)




I3 = IL32 + IS32 + IL33 + IS33 (314)



we know that

|x2| = max(r1 r2) =1

2(r1 + r2 + |r1 minus r2|)

|x3| = min(r3 r4) =1

2(r3 + r4 minus |r3 minus r4|) (315)



)(316)





I4 =sumlimi

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)


int |x4||x3|


=sum

limille0

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)

|x4|2


l minus 2



I43 =sum

limille0

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)

|x4|2

(minus|x3|lminus2

l minus 2


I44 =sum

limille0

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)

|x4|2

(|x4|lminus2

l minus 2

)(318)

ndash 12 ndash

JHEP03(2016)165




I4 = I43 + I44 (319)


I5 =sumlimi

Cm1m2m3m4

l1l2l3l4

prodi

Y limi(xi)

int infin|x4|


=sumlimil

Cm1m2m3m4

l1l2l3l4

prodi

Y limi(xi)

(|x4|minus4minusl

4 + l

)δ(l l1 + l2 + l3 + l4) (320)






C2 = ∆(∆minus 4) + l(l + 2) = minus3 (321)










and Libine [29]


result is

log(r3r2

)sumlimi

(Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

)δ(l1 + l2 l3 + l4) (41)


Φ(s t) = F0(s t) + log(s)F1(s t) (42)



ndash 13 ndash

JHEP03(2016)165





)spart

parts

(spartH

parts

)minus(


(tpartH

partt

)+ 4s

partH

parts(44)


C2 sΦ = 2 sΦ C2 sF1 = 2 sF1 (45)



for the integral is

1

2x213x224

F1(s t) (46)



plies that

2sumlimi

(Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

)δ(l1 + l2 l3 + l4) =

1

x213x224

F1(s t)

(47)




V+ otimes V+ = A[200] +

infinoplusk=1

D[k+2 k2 k2] (48)








P++++ defined by




ndash 14 ndash

JHEP03(2016)165



















1

2x213x224



prime24 P(x1 x2 x










ndash 15 ndash

JHEP03(2016)165





m1(x)Y l2

m2(x)Y l5

m5(x) (416)


(x1)Yl2m2


Y l1m1

(x1)otimes Y l2m2


(x)Y l2m2

(x)

Y l3m3

(xprime3)

|x3|2otimesY l4m4

(xprime4)

|x4|2rarr

Y l3m3

(xprime)

|x|2otimesY l4m4

(xprime)

|x|2(417)















m(K)Y ml (xprime)

sumlimi

(minusi)l12l1 l1

Y l1m1


Y l2m2

(x2)(i)l3

2l3 l3Y l3m3

(xprime3)(i)l4

2l4 l4Y l4m4

(xprime4)

P++minusminus

(Y m1l1




(K)vminus)

=sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

P++minusminus

(Y m1l1

(P )

2l1 l1v+ otimes

Y m2l2

(P )

2l2 l2v+ otimes

Y m3l3

(K)

2l3 l3vminus otimes

Y m4l4

(K)

2l4 l4vminus)

(418)








ndash 16 ndash

JHEP03(2016)165






F1 =1

λlog

((1 + ρs)


)(419)



sk

u2k+1[Qk(u) log(1 + u)]+ (420)



Qk(u) =

ksuml=0

bl(k)ul (421)

where

bl(k) =

(k

k minus l

)(2k + l

k minus l

)(422)








equivariant maps


I = I1 + I2 + I3 + I4 + I5



I(1) = (IS11 + IS21)

I(2) = (IS22 + IS32 + IL32)

I(3) = (IL33 + IS33 + IS43)

I(4) = (IS45 + IS55) (52)

ndash 17 ndash

JHEP03(2016)165




I = I(1) + I(2) + I(3) + I(4) (53)



jI(i) = minus2δij

prodk 6=j

1

(xk minus xi)2(54)









I(1) =sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

r22

Y l3m3

(xprime3)

r23

Y l4m4

(xprime4)



(l1 + l2 + l3 + l4 + 4)(minusl1 + l2 + l3 + l4 + 2)(55)




part2


(r21)A = 4A(A+ 1)(r21)Aminus1

part


part

partxmicro1Y l1m1

(x1)part


21)Aminus1Y l1

m1(x1) (56)


part2


I(1) =sumlimi


r22r23r

24

Y l1m1

(x1)Yl2m2

(xprime2)Yl3m3

(xprime3)Yl4m4


(57)


1


1

r22r23r

24

sumlimi

Y l2m2

(x1)Yl3m3

(x1)Yl4m4

(x1)Yl2m2

(xprime2)Yl3m3

(xprime3)Yl4m4

(xprime4)

=(l1 + 1)rl2+l3+l41

r22r23r

24

sumlimi

Cm1m2m3m4l1l2l3l4

Y l1m1

(x)Y l2m2

(xprime2)Yl3m3

(xprime3)Yl4m4

(xprime4) (58)

ndash 18 ndash

JHEP03(2016)165














I(2) =sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


1

l2+l3+l4minusl1+2

δ(l1 + l2 l3 + l4)

r23r24

+sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


1

r23r24

times



]


part

partxmicro2Y l2m2

(x2)part


22)Aminus1Y l2

m2(x2)

part

partxmicro2logr2 =

xmicro2r22

part

partxmicro2

part

partxmicro2logr2 =

2

r22(59)

We find

part2




r23r24

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


(510)


minus 2







ndash 19 ndash

JHEP03(2016)165


equivariant map


such that



x22x23x

24








23x

24


prime2)f(x1 x

prime3)f(x1 x

prime4)

(515)



1


(516)







part2



prime2)f(x1 x

prime3)f(x1 x

prime4) (518)


24I

(1)(x1 x2 x3 x4)


x22x23x

24 I(1)(x1 x2 x3 x4)rarr F1(e


While x23x24I








ndash 20 ndash

JHEP03(2016)165

implies that (part

partxmicro1+ x22

part

partxmicro2

1

x22+ x23

part

partxmicro3

1

x23+ x24

part

partxmicro4

1

x24

)F1 = 0 (521)


24I



24(I

(1)+I(2)+I(3)+I(4)) = x22x23x



partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)I = 0 (522)





23x

24I(x1 x2 x3 x4) =

x22x23x

24

x213x224

Φ(s t)

(523)



with


minus)=xprime21 x

prime22 xprime23


The x21x22x

23I













〈φ(x1)φ3(x2)φ(x3)φ(x4)〉 (528)




ndash 21 ndash

JHEP03(2016)165



twisted co-product










23x

24



















prime22 xprime23






ndash 22 ndash

JHEP03(2016)165





I2 =

intd4x5d

4x6xminus256

x215x225x

245x

226x

246x

236

(61)






















ndash 23 ndash

JHEP03(2016)165
































ndash 24 ndash

JHEP03(2016)165































Acknowledgments








this work was done

ndash 25 ndash

JHEP03(2016)165








1

2π2

intdS3Y l1

m1(x)Y l2

m2(x) = gm1m2

δl1l2

(l1 + 1)(A1)




gm1m2 = δm1m2


m2(x) = Y l

m1(x)

Y ml (x) = Y l

m(x) (A2)






(mLmR)

gm1m2 = δm1minusm2

Y ml (x) = Y l

minusm(x) (A3)



ables are

z = x1 + ix2 harr(

1

2

1

2


(minus1

2minus1

2

)w = x3 + ix4 harr

(1

2minus1

2


(minus1

21

2

)(A4)

We find that

Y l(l2l2) = (x1 + ix2)

l = zl (A5)

has normalization

1

2π2

intdS3(Y l

( l2 l2))lowast Y l

( l2 l2)

=1

(l + 1)(A6)


ndash 26 ndash

JHEP03(2016)165




(minusi)l

2llY lm(x)Y m

l (P )v+ (A7)


η(Y l1m1

(P )v+ Y m2l2

(K)vminus)

= δl1l2δm1m222l1(l1)2 (A8)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)= δl1l2δm1m2 (A9)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)=

1

2π2

intS3

Y m2l2

(xprime)


m1(x)) (A10)





Y lm(xprime)

xprime2


m(x) (A12)


η = f η


1


summ

1

y2Y lm(x)Y m


=infinsuml=0

summ

1

x2Y lm(xprime)Y m

l (y) if |x| gt |y| (A13)






ndash 27 ndash

JHEP03(2016)165


C2 = minus1

2LMNL


2MpqMqp (A14)






(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)eminusiP middotxv+

=

(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)ipart





Mpq = xppart

partxqminus xq

part

partxp(A16)


1

2MpqMqp(r

AY lm(x)) =

rA

2MpqMqpY


m(x) (A17)





m(x) (A19)

Thus we have

C2 rAY l




m(x) (A21)





F1 =1

λlog

((1 + ρs)


)(B1)

ndash 28 ndash

JHEP03(2016)165


of s to obtain

F1 =log(1+u)

u+

s


s2

u5(3u(2+u)minus 6 + 6u+ u2) log(1+u)

)+s3


)(B2)


F1 =

(minus1 +

u

2minus u2

3+u3


)+ s

(minus1

6+u

6minus 3u2


)+ s2

(minus1

30+

u

20minus 2u2


)+s3

(minus1

140+

u

70minus 5u2



The term at sk is

sk









in Pk(u) is

(2k)

(k)2(B6)


Qk(u) =ksuml=0

blul (B7)


ksuml=0

bl(minus1)l

(iminus l)= 0 (B8)




b1 =5b02 b2 =

20b09 b3 =

5b06 b4 =

5b042 b5 =

b0252

(B9)

ndash 29 ndash

JHEP03(2016)165



ksuml=0

MIlbl = 0 (B10)

where

MIl =(minus1)l




MIlbl (B12)



bl =sumI




bIb0 = 1 16bIb0 = 32 35 120bIb0 = 2 97 27 170

bIb0 =

5

2

20

95

6

5

42

1

252

bIb0 =

3

75

22

20

11

5

11

1

22

1

924

bIb0 =

7

2

63

13175

52

175

143

63

286

7

429

1

3432

(B14)



b1(k)b0 = k2

bk(k)b0 =k2

(2k)

bkminus1(k)b0 =k2

(2k)k(k + 1)

bkminus2(k)b0 =k2

(2k)

(k minus 1)k(k + 1)(k + 2)

4

bkminus3(k)b0 =k2

(2k)

(k

3

)(k + 3

3

)bkminusm(k)b0 =

k2

(2k)

(k

m

)(k +m

m

)(B15)

ndash 30 ndash

JHEP03(2016)165


so that

we have

b1(k) = k2(2k)

k2

bk(k) = 1



4

bkminus3(k) =

(k

3

)(k + 3

3

)bkminusm(k) =

(k

m

)(k +m

m

)(B16)


Pk(u) =

ksumi=1

aiuiminus1 (B17)



bl(minus1)iminusl

(iminus l)= 0 (B18)




k = 2 ~a = a1 ak = 6 3

k = 3 ~a =1

3times 60 60 11

k = 4 ~a =5

6times 84 126 52 5

k = 5 ~a =1

30times 7560 15120 9870 2310 137

k = 6 ~a =7

10times 13720 3300 2960 1140 174 7 (B19)

which leads to

a1(k) =(2k)

(k)2

ak(k) = 2h(k) = 2

ksuml=1

1

l(B20)




ndash 31 ndash

JHEP03(2016)165


grals






infinsumli=0

summi

infinsumt=0

sump


(x1)Yl2minusm2

(x2)Yl3m3

(xprime3)Yl4m4



t+ 1

=x23x

24

x213x224

1

λlog

((1 + ρs)


)(B21)



C l1 l2 lm1m2 m1+m2

=

langl12

l22

mL12

mL22

∣∣∣∣∣ l2

mL1 +mL2

2

ranglangl12

l22

mR12

mR22

∣∣∣∣∣ l2

mR1 +mR22






Yn12

n12n12

(x1)Yn22

n22n22



(x4) = zn11 zn2

2 zn33 zn4

4 (B22)


zn11 zn2

2 zn33 zn4


note that





that zn11 zn2

2 zn33 zn4


Yn12

n12n1minus2

2

(x1)Yn22

n2minus22

n22

(x2)Yn32


2

(x3)Yn42

minusn4minus22

minusn42

(x4)

=radicn1n2n3n4z

n1minus11 w1z

n2minus12 w2z

n3minus13 w3z

n4minus14 w4 (B24)


coefficients

Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B25)

Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B26)

ndash 32 ndash

JHEP03(2016)165


Cn12n22n1+n2

2n12n2minus2

2n1+n2minus2

2

=

radicn2

n1 + n2(B27)

Cn12n22n1+n2minus2

2n12n2minus2

2n1+n2minus2

2

= minusradic

n1n1 + n2

(B28)



n2minus12 w2z

n3minus13 w3z

n4minus14 w4 is

radicn1n2n3n4

[Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2 + 1

+Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2

]=


(B29)




section 5






parth








prime4vminus) (C1)



(2ixprime3microx

prime3 middot

part


part


)eiKmiddotx

prime3vminus

= ix23part

partxmicro3

1

x23eiKmiddotx

prime3vminus (C2)


parth


part

partxmicro2

(1

x22h

)minus x23

part

partxmicro3

(1

x23h

)minus x24

part

partxmicro4

(1

x24h

)(C3)

ndash 33 ndash

JHEP03(2016)165


x22x23x

24

(part

partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)[h

x22x23x

24

]= 0 (C4)


x1micropart

partx1νminus x1ν

part

partx1micro

)h= F1(Mmicroνe









x22x23x

24

(x1micro

part

partx1νminus x1ν

part


part

partx2νminus x2ν

part


part

partx3νminus x3ν

part

partx3micro

+x4micropart

partx4νminus x4ν

part

partx4micro

)[h

x22x23x

24

]= 0 (C6)


consider(x1 middot

part

partx1+ 3

)h = F1(e






x22x23x

24

(x1 middot

part

partx1+ x2 middot

part

partx2+ x3 middot

part

partx3+ x4 middot

part

partx4+ 12

)[h

x22x23x

24

]= 0 (C8)


h

x22x23x

24

=A




we have




ndash 34 ndash

JHEP03(2016)165





infinsumnm=0

im(minus1)n




where







minus (C14)

we find


and

infinsumn=0

(minus1)n




1


(C16)



=infinsum

n1n3n4=0

1



slot we get

α(P middot x2) =1


ndash 35 ndash

JHEP03(2016)165




=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)


=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)

(n1 + n3 + n4)



=infinsumn1

(minus1)n1




timesinfinsumn3

(minus1)n3




timesinfinsumn4

(minus1)n4





=1


1


1


=x22x

33x

24






References














ndash 36 ndash

JHEP03(2016)165










(2003) 013 [hep-th0212208] [INSPIRE]




(2014) 302 [arXiv14036646] [INSPIRE]






















(2015) 006 [arXiv150208005] [INSPIRE]






ndash 37 ndash

JHEP03(2016)165



[INSPIRE]

















(2015) 29FT01 [arXiv150500963] [INSPIRE]








[INSPIRE]














ndash 38 ndash

JHEP03(2016)165















ndash 39 ndash

Introduction























JHEP03(2016)165

where d3x = 12π2dS



write it as

Cm1m2m3m4

l1l2l3l4=

suml5m5m6

Cm1m2l5l1l2m5

1

(l5 + 1)Cm3m4l5l3l4m6


(32)

where

Y m1l1

Y m2l2

=suml5m5

Cm1m2l5l1l2m5

Y m5l5

(33)




or equivalently












I1 =sumlimi

prodi Y

limi(x

primei)

|x1|2|x2|2|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

|x1|4+l1+l2+l3+l44 + l1 + l2 + l3 + l4

(38)




ndash 10 ndash

JHEP03(2016)165


region is

I2 =sumlimi

Y l1m1

(x1)prodi=2 Y

limi(x

primei)

|x2|2|x3|2|x4|2

int |x2||x1|


l1(xprime)Y m2

l2(x)Y m3

l3(x)Y m4

l4(x)

=sumlimi

Y l1m1

(x1)prodi=2 Y

limi(x

primei)

|x2|2|x3|2|x4|2

int |x2||x1|


l1l2l3l4

=sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

|x2|2Y l3m3

(xprime3)

|x3|2Y l4m4

(xprime4)

|x4|2Cm1m2m3m4

l1l2l3l4

(minusl1 + l2 + l3 + l4 + 2)


1

)(39)





IS21 =sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

|x2|2Y l3m3

(xprime3)

|x3|2Y l4m4

(xprime4)

|x4|2Cm1m2m3m4

l1l2l3l4

(minusl1 + l2 + l3 + l4 + 2)


1

)IS22 =

sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

|x2|2Y l3m3

(xprime3)

|x3|2Y l4m4

(xprime4)

|x4|2Cm1m2m3m4

l1l2l3l4

(minusl1 + l2 + l3 + l4 + 2)


)(310)




primemicro3 x



I2 = IS21 + IR21 (311)


I3 =sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

int |x3||x2|


=sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4log

(|x3||x2|

)δ(l1 + l2 l3 + l4)

+sum

limil 6=0

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2

(|x3|l minus |x2|l

l


(312)

ndash 11 ndash

JHEP03(2016)165


IL32 =minussumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4log (|x2|) δ(l1 + l2 l3 + l4)

IS32 =sum

limil 6=0

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

(minus|x2|l

l


IL33 =sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4log (|x3|) δ(l1 + l2 l3 + l4)

IS33 =sum

limil 6=0

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

(|x3|l

l


(313)




I3 = IL32 + IS32 + IL33 + IS33 (314)



we know that

|x2| = max(r1 r2) =1

2(r1 + r2 + |r1 minus r2|)

|x3| = min(r3 r4) =1

2(r3 + r4 minus |r3 minus r4|) (315)



)(316)





I4 =sumlimi

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)


int |x4||x3|


=sum

limille0

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)

|x4|2


l minus 2



I43 =sum

limille0

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)

|x4|2

(minus|x3|lminus2

l minus 2


I44 =sum

limille0

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)

|x4|2

(|x4|lminus2

l minus 2

)(318)

ndash 12 ndash

JHEP03(2016)165




I4 = I43 + I44 (319)


I5 =sumlimi

Cm1m2m3m4

l1l2l3l4

prodi

Y limi(xi)

int infin|x4|


=sumlimil

Cm1m2m3m4

l1l2l3l4

prodi

Y limi(xi)

(|x4|minus4minusl

4 + l

)δ(l l1 + l2 + l3 + l4) (320)






C2 = ∆(∆minus 4) + l(l + 2) = minus3 (321)










and Libine [29]


result is

log(r3r2

)sumlimi

(Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

)δ(l1 + l2 l3 + l4) (41)


Φ(s t) = F0(s t) + log(s)F1(s t) (42)



ndash 13 ndash

JHEP03(2016)165





)spart

parts

(spartH

parts

)minus(


(tpartH

partt

)+ 4s

partH

parts(44)


C2 sΦ = 2 sΦ C2 sF1 = 2 sF1 (45)



for the integral is

1

2x213x224

F1(s t) (46)



plies that

2sumlimi

(Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

)δ(l1 + l2 l3 + l4) =

1

x213x224

F1(s t)

(47)




V+ otimes V+ = A[200] +

infinoplusk=1

D[k+2 k2 k2] (48)








P++++ defined by




ndash 14 ndash

JHEP03(2016)165



















1

2x213x224



prime24 P(x1 x2 x










ndash 15 ndash

JHEP03(2016)165





m1(x)Y l2

m2(x)Y l5

m5(x) (416)


(x1)Yl2m2


Y l1m1

(x1)otimes Y l2m2


(x)Y l2m2

(x)

Y l3m3

(xprime3)

|x3|2otimesY l4m4

(xprime4)

|x4|2rarr

Y l3m3

(xprime)

|x|2otimesY l4m4

(xprime)

|x|2(417)















m(K)Y ml (xprime)

sumlimi

(minusi)l12l1 l1

Y l1m1


Y l2m2

(x2)(i)l3

2l3 l3Y l3m3

(xprime3)(i)l4

2l4 l4Y l4m4

(xprime4)

P++minusminus

(Y m1l1




(K)vminus)

=sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

P++minusminus

(Y m1l1

(P )

2l1 l1v+ otimes

Y m2l2

(P )

2l2 l2v+ otimes

Y m3l3

(K)

2l3 l3vminus otimes

Y m4l4

(K)

2l4 l4vminus)

(418)








ndash 16 ndash

JHEP03(2016)165






F1 =1

λlog

((1 + ρs)


)(419)



sk

u2k+1[Qk(u) log(1 + u)]+ (420)



Qk(u) =

ksuml=0

bl(k)ul (421)

where

bl(k) =

(k

k minus l

)(2k + l

k minus l

)(422)








equivariant maps


I = I1 + I2 + I3 + I4 + I5



I(1) = (IS11 + IS21)

I(2) = (IS22 + IS32 + IL32)

I(3) = (IL33 + IS33 + IS43)

I(4) = (IS45 + IS55) (52)

ndash 17 ndash

JHEP03(2016)165




I = I(1) + I(2) + I(3) + I(4) (53)



jI(i) = minus2δij

prodk 6=j

1

(xk minus xi)2(54)









I(1) =sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

r22

Y l3m3

(xprime3)

r23

Y l4m4

(xprime4)



(l1 + l2 + l3 + l4 + 4)(minusl1 + l2 + l3 + l4 + 2)(55)




part2


(r21)A = 4A(A+ 1)(r21)Aminus1

part


part

partxmicro1Y l1m1

(x1)part


21)Aminus1Y l1

m1(x1) (56)


part2


I(1) =sumlimi


r22r23r

24

Y l1m1

(x1)Yl2m2

(xprime2)Yl3m3

(xprime3)Yl4m4


(57)


1


1

r22r23r

24

sumlimi

Y l2m2

(x1)Yl3m3

(x1)Yl4m4

(x1)Yl2m2

(xprime2)Yl3m3

(xprime3)Yl4m4

(xprime4)

=(l1 + 1)rl2+l3+l41

r22r23r

24

sumlimi

Cm1m2m3m4l1l2l3l4

Y l1m1

(x)Y l2m2

(xprime2)Yl3m3

(xprime3)Yl4m4

(xprime4) (58)

ndash 18 ndash

JHEP03(2016)165














I(2) =sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


1

l2+l3+l4minusl1+2

δ(l1 + l2 l3 + l4)

r23r24

+sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


1

r23r24

times



]


part

partxmicro2Y l2m2

(x2)part


22)Aminus1Y l2

m2(x2)

part

partxmicro2logr2 =

xmicro2r22

part

partxmicro2

part

partxmicro2logr2 =

2

r22(59)

We find

part2




r23r24

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


(510)


minus 2







ndash 19 ndash

JHEP03(2016)165


equivariant map


such that



x22x23x

24








23x

24


prime2)f(x1 x

prime3)f(x1 x

prime4)

(515)



1


(516)







part2



prime2)f(x1 x

prime3)f(x1 x

prime4) (518)


24I

(1)(x1 x2 x3 x4)


x22x23x

24 I(1)(x1 x2 x3 x4)rarr F1(e


While x23x24I








ndash 20 ndash

JHEP03(2016)165

implies that (part

partxmicro1+ x22

part

partxmicro2

1

x22+ x23

part

partxmicro3

1

x23+ x24

part

partxmicro4

1

x24

)F1 = 0 (521)


24I



24(I

(1)+I(2)+I(3)+I(4)) = x22x23x



partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)I = 0 (522)





23x

24I(x1 x2 x3 x4) =

x22x23x

24

x213x224

Φ(s t)

(523)



with


minus)=xprime21 x

prime22 xprime23


The x21x22x

23I













〈φ(x1)φ3(x2)φ(x3)φ(x4)〉 (528)




ndash 21 ndash

JHEP03(2016)165



twisted co-product










23x

24



















prime22 xprime23






ndash 22 ndash

JHEP03(2016)165





I2 =

intd4x5d

4x6xminus256

x215x225x

245x

226x

246x

236

(61)






















ndash 23 ndash

JHEP03(2016)165
































ndash 24 ndash

JHEP03(2016)165































Acknowledgments








this work was done

ndash 25 ndash

JHEP03(2016)165








1

2π2

intdS3Y l1

m1(x)Y l2

m2(x) = gm1m2

δl1l2

(l1 + 1)(A1)




gm1m2 = δm1m2


m2(x) = Y l

m1(x)

Y ml (x) = Y l

m(x) (A2)






(mLmR)

gm1m2 = δm1minusm2

Y ml (x) = Y l

minusm(x) (A3)



ables are

z = x1 + ix2 harr(

1

2

1

2


(minus1

2minus1

2

)w = x3 + ix4 harr

(1

2minus1

2


(minus1

21

2

)(A4)

We find that

Y l(l2l2) = (x1 + ix2)

l = zl (A5)

has normalization

1

2π2

intdS3(Y l

( l2 l2))lowast Y l

( l2 l2)

=1

(l + 1)(A6)


ndash 26 ndash

JHEP03(2016)165




(minusi)l

2llY lm(x)Y m

l (P )v+ (A7)


η(Y l1m1

(P )v+ Y m2l2

(K)vminus)

= δl1l2δm1m222l1(l1)2 (A8)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)= δl1l2δm1m2 (A9)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)=

1

2π2

intS3

Y m2l2

(xprime)


m1(x)) (A10)





Y lm(xprime)

xprime2


m(x) (A12)


η = f η


1


summ

1

y2Y lm(x)Y m


=infinsuml=0

summ

1

x2Y lm(xprime)Y m

l (y) if |x| gt |y| (A13)






ndash 27 ndash

JHEP03(2016)165


C2 = minus1

2LMNL


2MpqMqp (A14)






(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)eminusiP middotxv+

=

(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)ipart





Mpq = xppart

partxqminus xq

part

partxp(A16)


1

2MpqMqp(r

AY lm(x)) =

rA

2MpqMqpY


m(x) (A17)





m(x) (A19)

Thus we have

C2 rAY l




m(x) (A21)





F1 =1

λlog

((1 + ρs)


)(B1)

ndash 28 ndash

JHEP03(2016)165


of s to obtain

F1 =log(1+u)

u+

s


s2

u5(3u(2+u)minus 6 + 6u+ u2) log(1+u)

)+s3


)(B2)


F1 =

(minus1 +

u

2minus u2

3+u3


)+ s

(minus1

6+u

6minus 3u2


)+ s2

(minus1

30+

u

20minus 2u2


)+s3

(minus1

140+

u

70minus 5u2



The term at sk is

sk









in Pk(u) is

(2k)

(k)2(B6)


Qk(u) =ksuml=0

blul (B7)


ksuml=0

bl(minus1)l

(iminus l)= 0 (B8)




b1 =5b02 b2 =

20b09 b3 =

5b06 b4 =

5b042 b5 =

b0252

(B9)

ndash 29 ndash

JHEP03(2016)165



ksuml=0

MIlbl = 0 (B10)

where

MIl =(minus1)l




MIlbl (B12)



bl =sumI




bIb0 = 1 16bIb0 = 32 35 120bIb0 = 2 97 27 170

bIb0 =

5

2

20

95

6

5

42

1

252

bIb0 =

3

75

22

20

11

5

11

1

22

1

924

bIb0 =

7

2

63

13175

52

175

143

63

286

7

429

1

3432

(B14)



b1(k)b0 = k2

bk(k)b0 =k2

(2k)

bkminus1(k)b0 =k2

(2k)k(k + 1)

bkminus2(k)b0 =k2

(2k)

(k minus 1)k(k + 1)(k + 2)

4

bkminus3(k)b0 =k2

(2k)

(k

3

)(k + 3

3

)bkminusm(k)b0 =

k2

(2k)

(k

m

)(k +m

m

)(B15)

ndash 30 ndash

JHEP03(2016)165


so that

we have

b1(k) = k2(2k)

k2

bk(k) = 1



4

bkminus3(k) =

(k

3

)(k + 3

3

)bkminusm(k) =

(k

m

)(k +m

m

)(B16)


Pk(u) =

ksumi=1

aiuiminus1 (B17)



bl(minus1)iminusl

(iminus l)= 0 (B18)




k = 2 ~a = a1 ak = 6 3

k = 3 ~a =1

3times 60 60 11

k = 4 ~a =5

6times 84 126 52 5

k = 5 ~a =1

30times 7560 15120 9870 2310 137

k = 6 ~a =7

10times 13720 3300 2960 1140 174 7 (B19)

which leads to

a1(k) =(2k)

(k)2

ak(k) = 2h(k) = 2

ksuml=1

1

l(B20)




ndash 31 ndash

JHEP03(2016)165


grals






infinsumli=0

summi

infinsumt=0

sump


(x1)Yl2minusm2

(x2)Yl3m3

(xprime3)Yl4m4



t+ 1

=x23x

24

x213x224

1

λlog

((1 + ρs)


)(B21)



C l1 l2 lm1m2 m1+m2

=

langl12

l22

mL12

mL22

∣∣∣∣∣ l2

mL1 +mL2

2

ranglangl12

l22

mR12

mR22

∣∣∣∣∣ l2

mR1 +mR22






Yn12

n12n12

(x1)Yn22

n22n22



(x4) = zn11 zn2

2 zn33 zn4

4 (B22)


zn11 zn2

2 zn33 zn4


note that





that zn11 zn2

2 zn33 zn4


Yn12

n12n1minus2

2

(x1)Yn22

n2minus22

n22

(x2)Yn32


2

(x3)Yn42

minusn4minus22

minusn42

(x4)

=radicn1n2n3n4z

n1minus11 w1z

n2minus12 w2z

n3minus13 w3z

n4minus14 w4 (B24)


coefficients

Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B25)

Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B26)

ndash 32 ndash

JHEP03(2016)165


Cn12n22n1+n2

2n12n2minus2

2n1+n2minus2

2

=

radicn2

n1 + n2(B27)

Cn12n22n1+n2minus2

2n12n2minus2

2n1+n2minus2

2

= minusradic

n1n1 + n2

(B28)



n2minus12 w2z

n3minus13 w3z

n4minus14 w4 is

radicn1n2n3n4

[Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2 + 1

+Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2

]=


(B29)




section 5






parth








prime4vminus) (C1)



(2ixprime3microx

prime3 middot

part


part


)eiKmiddotx

prime3vminus

= ix23part

partxmicro3

1

x23eiKmiddotx

prime3vminus (C2)


parth


part

partxmicro2

(1

x22h

)minus x23

part

partxmicro3

(1

x23h

)minus x24

part

partxmicro4

(1

x24h

)(C3)

ndash 33 ndash

JHEP03(2016)165


x22x23x

24

(part

partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)[h

x22x23x

24

]= 0 (C4)


x1micropart

partx1νminus x1ν

part

partx1micro

)h= F1(Mmicroνe









x22x23x

24

(x1micro

part

partx1νminus x1ν

part


part

partx2νminus x2ν

part


part

partx3νminus x3ν

part

partx3micro

+x4micropart

partx4νminus x4ν

part

partx4micro

)[h

x22x23x

24

]= 0 (C6)


consider(x1 middot

part

partx1+ 3

)h = F1(e






x22x23x

24

(x1 middot

part

partx1+ x2 middot

part

partx2+ x3 middot

part

partx3+ x4 middot

part

partx4+ 12

)[h

x22x23x

24

]= 0 (C8)


h

x22x23x

24

=A




we have




ndash 34 ndash

JHEP03(2016)165





infinsumnm=0

im(minus1)n




where







minus (C14)

we find


and

infinsumn=0

(minus1)n




1


(C16)



=infinsum

n1n3n4=0

1



slot we get

α(P middot x2) =1


ndash 35 ndash

JHEP03(2016)165




=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)


=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)

(n1 + n3 + n4)



=infinsumn1

(minus1)n1




timesinfinsumn3

(minus1)n3




timesinfinsumn4

(minus1)n4





=1


1


1


=x22x

33x

24






References














ndash 36 ndash

JHEP03(2016)165










(2003) 013 [hep-th0212208] [INSPIRE]




(2014) 302 [arXiv14036646] [INSPIRE]






















(2015) 006 [arXiv150208005] [INSPIRE]






ndash 37 ndash

JHEP03(2016)165



[INSPIRE]

















(2015) 29FT01 [arXiv150500963] [INSPIRE]








[INSPIRE]














ndash 38 ndash

JHEP03(2016)165















ndash 39 ndash

Introduction























JHEP03(2016)165


region is

I2 =sumlimi

Y l1m1

(x1)prodi=2 Y

limi(x

primei)

|x2|2|x3|2|x4|2

int |x2||x1|


l1(xprime)Y m2

l2(x)Y m3

l3(x)Y m4

l4(x)

=sumlimi

Y l1m1

(x1)prodi=2 Y

limi(x

primei)

|x2|2|x3|2|x4|2

int |x2||x1|


l1l2l3l4

=sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

|x2|2Y l3m3

(xprime3)

|x3|2Y l4m4

(xprime4)

|x4|2Cm1m2m3m4

l1l2l3l4

(minusl1 + l2 + l3 + l4 + 2)


1

)(39)





IS21 =sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

|x2|2Y l3m3

(xprime3)

|x3|2Y l4m4

(xprime4)

|x4|2Cm1m2m3m4

l1l2l3l4

(minusl1 + l2 + l3 + l4 + 2)


1

)IS22 =

sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

|x2|2Y l3m3

(xprime3)

|x3|2Y l4m4

(xprime4)

|x4|2Cm1m2m3m4

l1l2l3l4

(minusl1 + l2 + l3 + l4 + 2)


)(310)




primemicro3 x



I2 = IS21 + IR21 (311)


I3 =sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

int |x3||x2|


=sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4log

(|x3||x2|

)δ(l1 + l2 l3 + l4)

+sum

limil 6=0

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2

(|x3|l minus |x2|l

l


(312)

ndash 11 ndash

JHEP03(2016)165


IL32 =minussumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4log (|x2|) δ(l1 + l2 l3 + l4)

IS32 =sum

limil 6=0

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

(minus|x2|l

l


IL33 =sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4log (|x3|) δ(l1 + l2 l3 + l4)

IS33 =sum

limil 6=0

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

(|x3|l

l


(313)




I3 = IL32 + IS32 + IL33 + IS33 (314)



we know that

|x2| = max(r1 r2) =1

2(r1 + r2 + |r1 minus r2|)

|x3| = min(r3 r4) =1

2(r3 + r4 minus |r3 minus r4|) (315)



)(316)





I4 =sumlimi

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)


int |x4||x3|


=sum

limille0

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)

|x4|2


l minus 2



I43 =sum

limille0

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)

|x4|2

(minus|x3|lminus2

l minus 2


I44 =sum

limille0

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)

|x4|2

(|x4|lminus2

l minus 2

)(318)

ndash 12 ndash

JHEP03(2016)165




I4 = I43 + I44 (319)


I5 =sumlimi

Cm1m2m3m4

l1l2l3l4

prodi

Y limi(xi)

int infin|x4|


=sumlimil

Cm1m2m3m4

l1l2l3l4

prodi

Y limi(xi)

(|x4|minus4minusl

4 + l

)δ(l l1 + l2 + l3 + l4) (320)






C2 = ∆(∆minus 4) + l(l + 2) = minus3 (321)










and Libine [29]


result is

log(r3r2

)sumlimi

(Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

)δ(l1 + l2 l3 + l4) (41)


Φ(s t) = F0(s t) + log(s)F1(s t) (42)



ndash 13 ndash

JHEP03(2016)165





)spart

parts

(spartH

parts

)minus(


(tpartH

partt

)+ 4s

partH

parts(44)


C2 sΦ = 2 sΦ C2 sF1 = 2 sF1 (45)



for the integral is

1

2x213x224

F1(s t) (46)



plies that

2sumlimi

(Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

)δ(l1 + l2 l3 + l4) =

1

x213x224

F1(s t)

(47)




V+ otimes V+ = A[200] +

infinoplusk=1

D[k+2 k2 k2] (48)








P++++ defined by




ndash 14 ndash

JHEP03(2016)165



















1

2x213x224



prime24 P(x1 x2 x










ndash 15 ndash

JHEP03(2016)165





m1(x)Y l2

m2(x)Y l5

m5(x) (416)


(x1)Yl2m2


Y l1m1

(x1)otimes Y l2m2


(x)Y l2m2

(x)

Y l3m3

(xprime3)

|x3|2otimesY l4m4

(xprime4)

|x4|2rarr

Y l3m3

(xprime)

|x|2otimesY l4m4

(xprime)

|x|2(417)















m(K)Y ml (xprime)

sumlimi

(minusi)l12l1 l1

Y l1m1


Y l2m2

(x2)(i)l3

2l3 l3Y l3m3

(xprime3)(i)l4

2l4 l4Y l4m4

(xprime4)

P++minusminus

(Y m1l1




(K)vminus)

=sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

P++minusminus

(Y m1l1

(P )

2l1 l1v+ otimes

Y m2l2

(P )

2l2 l2v+ otimes

Y m3l3

(K)

2l3 l3vminus otimes

Y m4l4

(K)

2l4 l4vminus)

(418)








ndash 16 ndash

JHEP03(2016)165






F1 =1

λlog

((1 + ρs)


)(419)



sk

u2k+1[Qk(u) log(1 + u)]+ (420)



Qk(u) =

ksuml=0

bl(k)ul (421)

where

bl(k) =

(k

k minus l

)(2k + l

k minus l

)(422)








equivariant maps


I = I1 + I2 + I3 + I4 + I5



I(1) = (IS11 + IS21)

I(2) = (IS22 + IS32 + IL32)

I(3) = (IL33 + IS33 + IS43)

I(4) = (IS45 + IS55) (52)

ndash 17 ndash

JHEP03(2016)165




I = I(1) + I(2) + I(3) + I(4) (53)



jI(i) = minus2δij

prodk 6=j

1

(xk minus xi)2(54)









I(1) =sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

r22

Y l3m3

(xprime3)

r23

Y l4m4

(xprime4)



(l1 + l2 + l3 + l4 + 4)(minusl1 + l2 + l3 + l4 + 2)(55)




part2


(r21)A = 4A(A+ 1)(r21)Aminus1

part


part

partxmicro1Y l1m1

(x1)part


21)Aminus1Y l1

m1(x1) (56)


part2


I(1) =sumlimi


r22r23r

24

Y l1m1

(x1)Yl2m2

(xprime2)Yl3m3

(xprime3)Yl4m4


(57)


1


1

r22r23r

24

sumlimi

Y l2m2

(x1)Yl3m3

(x1)Yl4m4

(x1)Yl2m2

(xprime2)Yl3m3

(xprime3)Yl4m4

(xprime4)

=(l1 + 1)rl2+l3+l41

r22r23r

24

sumlimi

Cm1m2m3m4l1l2l3l4

Y l1m1

(x)Y l2m2

(xprime2)Yl3m3

(xprime3)Yl4m4

(xprime4) (58)

ndash 18 ndash

JHEP03(2016)165














I(2) =sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


1

l2+l3+l4minusl1+2

δ(l1 + l2 l3 + l4)

r23r24

+sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


1

r23r24

times



]


part

partxmicro2Y l2m2

(x2)part


22)Aminus1Y l2

m2(x2)

part

partxmicro2logr2 =

xmicro2r22

part

partxmicro2

part

partxmicro2logr2 =

2

r22(59)

We find

part2




r23r24

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


(510)


minus 2







ndash 19 ndash

JHEP03(2016)165


equivariant map


such that



x22x23x

24








23x

24


prime2)f(x1 x

prime3)f(x1 x

prime4)

(515)



1


(516)







part2



prime2)f(x1 x

prime3)f(x1 x

prime4) (518)


24I

(1)(x1 x2 x3 x4)


x22x23x

24 I(1)(x1 x2 x3 x4)rarr F1(e


While x23x24I








ndash 20 ndash

JHEP03(2016)165

implies that (part

partxmicro1+ x22

part

partxmicro2

1

x22+ x23

part

partxmicro3

1

x23+ x24

part

partxmicro4

1

x24

)F1 = 0 (521)


24I



24(I

(1)+I(2)+I(3)+I(4)) = x22x23x



partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)I = 0 (522)





23x

24I(x1 x2 x3 x4) =

x22x23x

24

x213x224

Φ(s t)

(523)



with


minus)=xprime21 x

prime22 xprime23


The x21x22x

23I













〈φ(x1)φ3(x2)φ(x3)φ(x4)〉 (528)




ndash 21 ndash

JHEP03(2016)165



twisted co-product










23x

24



















prime22 xprime23






ndash 22 ndash

JHEP03(2016)165





I2 =

intd4x5d

4x6xminus256

x215x225x

245x

226x

246x

236

(61)






















ndash 23 ndash

JHEP03(2016)165
































ndash 24 ndash

JHEP03(2016)165































Acknowledgments








this work was done

ndash 25 ndash

JHEP03(2016)165








1

2π2

intdS3Y l1

m1(x)Y l2

m2(x) = gm1m2

δl1l2

(l1 + 1)(A1)




gm1m2 = δm1m2


m2(x) = Y l

m1(x)

Y ml (x) = Y l

m(x) (A2)






(mLmR)

gm1m2 = δm1minusm2

Y ml (x) = Y l

minusm(x) (A3)



ables are

z = x1 + ix2 harr(

1

2

1

2


(minus1

2minus1

2

)w = x3 + ix4 harr

(1

2minus1

2


(minus1

21

2

)(A4)

We find that

Y l(l2l2) = (x1 + ix2)

l = zl (A5)

has normalization

1

2π2

intdS3(Y l

( l2 l2))lowast Y l

( l2 l2)

=1

(l + 1)(A6)


ndash 26 ndash

JHEP03(2016)165




(minusi)l

2llY lm(x)Y m

l (P )v+ (A7)


η(Y l1m1

(P )v+ Y m2l2

(K)vminus)

= δl1l2δm1m222l1(l1)2 (A8)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)= δl1l2δm1m2 (A9)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)=

1

2π2

intS3

Y m2l2

(xprime)


m1(x)) (A10)





Y lm(xprime)

xprime2


m(x) (A12)


η = f η


1


summ

1

y2Y lm(x)Y m


=infinsuml=0

summ

1

x2Y lm(xprime)Y m

l (y) if |x| gt |y| (A13)






ndash 27 ndash

JHEP03(2016)165


C2 = minus1

2LMNL


2MpqMqp (A14)






(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)eminusiP middotxv+

=

(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)ipart





Mpq = xppart

partxqminus xq

part

partxp(A16)


1

2MpqMqp(r

AY lm(x)) =

rA

2MpqMqpY


m(x) (A17)





m(x) (A19)

Thus we have

C2 rAY l




m(x) (A21)





F1 =1

λlog

((1 + ρs)


)(B1)

ndash 28 ndash

JHEP03(2016)165


of s to obtain

F1 =log(1+u)

u+

s


s2

u5(3u(2+u)minus 6 + 6u+ u2) log(1+u)

)+s3


)(B2)


F1 =

(minus1 +

u

2minus u2

3+u3


)+ s

(minus1

6+u

6minus 3u2


)+ s2

(minus1

30+

u

20minus 2u2


)+s3

(minus1

140+

u

70minus 5u2



The term at sk is

sk









in Pk(u) is

(2k)

(k)2(B6)


Qk(u) =ksuml=0

blul (B7)


ksuml=0

bl(minus1)l

(iminus l)= 0 (B8)




b1 =5b02 b2 =

20b09 b3 =

5b06 b4 =

5b042 b5 =

b0252

(B9)

ndash 29 ndash

JHEP03(2016)165



ksuml=0

MIlbl = 0 (B10)

where

MIl =(minus1)l




MIlbl (B12)



bl =sumI




bIb0 = 1 16bIb0 = 32 35 120bIb0 = 2 97 27 170

bIb0 =

5

2

20

95

6

5

42

1

252

bIb0 =

3

75

22

20

11

5

11

1

22

1

924

bIb0 =

7

2

63

13175

52

175

143

63

286

7

429

1

3432

(B14)



b1(k)b0 = k2

bk(k)b0 =k2

(2k)

bkminus1(k)b0 =k2

(2k)k(k + 1)

bkminus2(k)b0 =k2

(2k)

(k minus 1)k(k + 1)(k + 2)

4

bkminus3(k)b0 =k2

(2k)

(k

3

)(k + 3

3

)bkminusm(k)b0 =

k2

(2k)

(k

m

)(k +m

m

)(B15)

ndash 30 ndash

JHEP03(2016)165


so that

we have

b1(k) = k2(2k)

k2

bk(k) = 1



4

bkminus3(k) =

(k

3

)(k + 3

3

)bkminusm(k) =

(k

m

)(k +m

m

)(B16)


Pk(u) =

ksumi=1

aiuiminus1 (B17)



bl(minus1)iminusl

(iminus l)= 0 (B18)




k = 2 ~a = a1 ak = 6 3

k = 3 ~a =1

3times 60 60 11

k = 4 ~a =5

6times 84 126 52 5

k = 5 ~a =1

30times 7560 15120 9870 2310 137

k = 6 ~a =7

10times 13720 3300 2960 1140 174 7 (B19)

which leads to

a1(k) =(2k)

(k)2

ak(k) = 2h(k) = 2

ksuml=1

1

l(B20)




ndash 31 ndash

JHEP03(2016)165


grals






infinsumli=0

summi

infinsumt=0

sump


(x1)Yl2minusm2

(x2)Yl3m3

(xprime3)Yl4m4



t+ 1

=x23x

24

x213x224

1

λlog

((1 + ρs)


)(B21)



C l1 l2 lm1m2 m1+m2

=

langl12

l22

mL12

mL22

∣∣∣∣∣ l2

mL1 +mL2

2

ranglangl12

l22

mR12

mR22

∣∣∣∣∣ l2

mR1 +mR22






Yn12

n12n12

(x1)Yn22

n22n22



(x4) = zn11 zn2

2 zn33 zn4

4 (B22)


zn11 zn2

2 zn33 zn4


note that





that zn11 zn2

2 zn33 zn4


Yn12

n12n1minus2

2

(x1)Yn22

n2minus22

n22

(x2)Yn32


2

(x3)Yn42

minusn4minus22

minusn42

(x4)

=radicn1n2n3n4z

n1minus11 w1z

n2minus12 w2z

n3minus13 w3z

n4minus14 w4 (B24)


coefficients

Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B25)

Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B26)

ndash 32 ndash

JHEP03(2016)165


Cn12n22n1+n2

2n12n2minus2

2n1+n2minus2

2

=

radicn2

n1 + n2(B27)

Cn12n22n1+n2minus2

2n12n2minus2

2n1+n2minus2

2

= minusradic

n1n1 + n2

(B28)



n2minus12 w2z

n3minus13 w3z

n4minus14 w4 is

radicn1n2n3n4

[Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2 + 1

+Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2

]=


(B29)




section 5






parth








prime4vminus) (C1)



(2ixprime3microx

prime3 middot

part


part


)eiKmiddotx

prime3vminus

= ix23part

partxmicro3

1

x23eiKmiddotx

prime3vminus (C2)


parth


part

partxmicro2

(1

x22h

)minus x23

part

partxmicro3

(1

x23h

)minus x24

part

partxmicro4

(1

x24h

)(C3)

ndash 33 ndash

JHEP03(2016)165


x22x23x

24

(part

partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)[h

x22x23x

24

]= 0 (C4)


x1micropart

partx1νminus x1ν

part

partx1micro

)h= F1(Mmicroνe









x22x23x

24

(x1micro

part

partx1νminus x1ν

part


part

partx2νminus x2ν

part


part

partx3νminus x3ν

part

partx3micro

+x4micropart

partx4νminus x4ν

part

partx4micro

)[h

x22x23x

24

]= 0 (C6)


consider(x1 middot

part

partx1+ 3

)h = F1(e






x22x23x

24

(x1 middot

part

partx1+ x2 middot

part

partx2+ x3 middot

part

partx3+ x4 middot

part

partx4+ 12

)[h

x22x23x

24

]= 0 (C8)


h

x22x23x

24

=A




we have




ndash 34 ndash

JHEP03(2016)165





infinsumnm=0

im(minus1)n




where







minus (C14)

we find


and

infinsumn=0

(minus1)n




1


(C16)



=infinsum

n1n3n4=0

1



slot we get

α(P middot x2) =1


ndash 35 ndash

JHEP03(2016)165




=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)


=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)

(n1 + n3 + n4)



=infinsumn1

(minus1)n1




timesinfinsumn3

(minus1)n3




timesinfinsumn4

(minus1)n4





=1


1


1


=x22x

33x

24






References














ndash 36 ndash

JHEP03(2016)165










(2003) 013 [hep-th0212208] [INSPIRE]




(2014) 302 [arXiv14036646] [INSPIRE]






















(2015) 006 [arXiv150208005] [INSPIRE]






ndash 37 ndash

JHEP03(2016)165



[INSPIRE]

















(2015) 29FT01 [arXiv150500963] [INSPIRE]








[INSPIRE]














ndash 38 ndash

JHEP03(2016)165















ndash 39 ndash

Introduction























JHEP03(2016)165


IL32 =minussumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4log (|x2|) δ(l1 + l2 l3 + l4)

IS32 =sum

limil 6=0

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

(minus|x2|l

l


IL33 =sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4log (|x3|) δ(l1 + l2 l3 + l4)

IS33 =sum

limil 6=0

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

(|x3|l

l


(313)




I3 = IL32 + IS32 + IL33 + IS33 (314)



we know that

|x2| = max(r1 r2) =1

2(r1 + r2 + |r1 minus r2|)

|x3| = min(r3 r4) =1

2(r3 + r4 minus |r3 minus r4|) (315)



)(316)





I4 =sumlimi

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)


int |x4||x3|


=sum

limille0

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)

|x4|2


l minus 2



I43 =sum

limille0

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)

|x4|2

(minus|x3|lminus2

l minus 2


I44 =sum

limille0

Y l4m4

(xprime4)prod3i=1 Y

limi(xi)

|x4|2

(|x4|lminus2

l minus 2

)(318)

ndash 12 ndash

JHEP03(2016)165




I4 = I43 + I44 (319)


I5 =sumlimi

Cm1m2m3m4

l1l2l3l4

prodi

Y limi(xi)

int infin|x4|


=sumlimil

Cm1m2m3m4

l1l2l3l4

prodi

Y limi(xi)

(|x4|minus4minusl

4 + l

)δ(l l1 + l2 + l3 + l4) (320)






C2 = ∆(∆minus 4) + l(l + 2) = minus3 (321)










and Libine [29]


result is

log(r3r2

)sumlimi

(Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

)δ(l1 + l2 l3 + l4) (41)


Φ(s t) = F0(s t) + log(s)F1(s t) (42)



ndash 13 ndash

JHEP03(2016)165





)spart

parts

(spartH

parts

)minus(


(tpartH

partt

)+ 4s

partH

parts(44)


C2 sΦ = 2 sΦ C2 sF1 = 2 sF1 (45)



for the integral is

1

2x213x224

F1(s t) (46)



plies that

2sumlimi

(Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

)δ(l1 + l2 l3 + l4) =

1

x213x224

F1(s t)

(47)




V+ otimes V+ = A[200] +

infinoplusk=1

D[k+2 k2 k2] (48)








P++++ defined by




ndash 14 ndash

JHEP03(2016)165



















1

2x213x224



prime24 P(x1 x2 x










ndash 15 ndash

JHEP03(2016)165





m1(x)Y l2

m2(x)Y l5

m5(x) (416)


(x1)Yl2m2


Y l1m1

(x1)otimes Y l2m2


(x)Y l2m2

(x)

Y l3m3

(xprime3)

|x3|2otimesY l4m4

(xprime4)

|x4|2rarr

Y l3m3

(xprime)

|x|2otimesY l4m4

(xprime)

|x|2(417)















m(K)Y ml (xprime)

sumlimi

(minusi)l12l1 l1

Y l1m1


Y l2m2

(x2)(i)l3

2l3 l3Y l3m3

(xprime3)(i)l4

2l4 l4Y l4m4

(xprime4)

P++minusminus

(Y m1l1




(K)vminus)

=sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

P++minusminus

(Y m1l1

(P )

2l1 l1v+ otimes

Y m2l2

(P )

2l2 l2v+ otimes

Y m3l3

(K)

2l3 l3vminus otimes

Y m4l4

(K)

2l4 l4vminus)

(418)








ndash 16 ndash

JHEP03(2016)165






F1 =1

λlog

((1 + ρs)


)(419)



sk

u2k+1[Qk(u) log(1 + u)]+ (420)



Qk(u) =

ksuml=0

bl(k)ul (421)

where

bl(k) =

(k

k minus l

)(2k + l

k minus l

)(422)








equivariant maps


I = I1 + I2 + I3 + I4 + I5



I(1) = (IS11 + IS21)

I(2) = (IS22 + IS32 + IL32)

I(3) = (IL33 + IS33 + IS43)

I(4) = (IS45 + IS55) (52)

ndash 17 ndash

JHEP03(2016)165




I = I(1) + I(2) + I(3) + I(4) (53)



jI(i) = minus2δij

prodk 6=j

1

(xk minus xi)2(54)









I(1) =sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

r22

Y l3m3

(xprime3)

r23

Y l4m4

(xprime4)



(l1 + l2 + l3 + l4 + 4)(minusl1 + l2 + l3 + l4 + 2)(55)




part2


(r21)A = 4A(A+ 1)(r21)Aminus1

part


part

partxmicro1Y l1m1

(x1)part


21)Aminus1Y l1

m1(x1) (56)


part2


I(1) =sumlimi


r22r23r

24

Y l1m1

(x1)Yl2m2

(xprime2)Yl3m3

(xprime3)Yl4m4


(57)


1


1

r22r23r

24

sumlimi

Y l2m2

(x1)Yl3m3

(x1)Yl4m4

(x1)Yl2m2

(xprime2)Yl3m3

(xprime3)Yl4m4

(xprime4)

=(l1 + 1)rl2+l3+l41

r22r23r

24

sumlimi

Cm1m2m3m4l1l2l3l4

Y l1m1

(x)Y l2m2

(xprime2)Yl3m3

(xprime3)Yl4m4

(xprime4) (58)

ndash 18 ndash

JHEP03(2016)165














I(2) =sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


1

l2+l3+l4minusl1+2

δ(l1 + l2 l3 + l4)

r23r24

+sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


1

r23r24

times



]


part

partxmicro2Y l2m2

(x2)part


22)Aminus1Y l2

m2(x2)

part

partxmicro2logr2 =

xmicro2r22

part

partxmicro2

part

partxmicro2logr2 =

2

r22(59)

We find

part2




r23r24

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


(510)


minus 2







ndash 19 ndash

JHEP03(2016)165


equivariant map


such that



x22x23x

24








23x

24


prime2)f(x1 x

prime3)f(x1 x

prime4)

(515)



1


(516)







part2



prime2)f(x1 x

prime3)f(x1 x

prime4) (518)


24I

(1)(x1 x2 x3 x4)


x22x23x

24 I(1)(x1 x2 x3 x4)rarr F1(e


While x23x24I








ndash 20 ndash

JHEP03(2016)165

implies that (part

partxmicro1+ x22

part

partxmicro2

1

x22+ x23

part

partxmicro3

1

x23+ x24

part

partxmicro4

1

x24

)F1 = 0 (521)


24I



24(I

(1)+I(2)+I(3)+I(4)) = x22x23x



partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)I = 0 (522)





23x

24I(x1 x2 x3 x4) =

x22x23x

24

x213x224

Φ(s t)

(523)



with


minus)=xprime21 x

prime22 xprime23


The x21x22x

23I













〈φ(x1)φ3(x2)φ(x3)φ(x4)〉 (528)




ndash 21 ndash

JHEP03(2016)165



twisted co-product










23x

24



















prime22 xprime23






ndash 22 ndash

JHEP03(2016)165





I2 =

intd4x5d

4x6xminus256

x215x225x

245x

226x

246x

236

(61)






















ndash 23 ndash

JHEP03(2016)165
































ndash 24 ndash

JHEP03(2016)165































Acknowledgments








this work was done

ndash 25 ndash

JHEP03(2016)165








1

2π2

intdS3Y l1

m1(x)Y l2

m2(x) = gm1m2

δl1l2

(l1 + 1)(A1)




gm1m2 = δm1m2


m2(x) = Y l

m1(x)

Y ml (x) = Y l

m(x) (A2)






(mLmR)

gm1m2 = δm1minusm2

Y ml (x) = Y l

minusm(x) (A3)



ables are

z = x1 + ix2 harr(

1

2

1

2


(minus1

2minus1

2

)w = x3 + ix4 harr

(1

2minus1

2


(minus1

21

2

)(A4)

We find that

Y l(l2l2) = (x1 + ix2)

l = zl (A5)

has normalization

1

2π2

intdS3(Y l

( l2 l2))lowast Y l

( l2 l2)

=1

(l + 1)(A6)


ndash 26 ndash

JHEP03(2016)165




(minusi)l

2llY lm(x)Y m

l (P )v+ (A7)


η(Y l1m1

(P )v+ Y m2l2

(K)vminus)

= δl1l2δm1m222l1(l1)2 (A8)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)= δl1l2δm1m2 (A9)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)=

1

2π2

intS3

Y m2l2

(xprime)


m1(x)) (A10)





Y lm(xprime)

xprime2


m(x) (A12)


η = f η


1


summ

1

y2Y lm(x)Y m


=infinsuml=0

summ

1

x2Y lm(xprime)Y m

l (y) if |x| gt |y| (A13)






ndash 27 ndash

JHEP03(2016)165


C2 = minus1

2LMNL


2MpqMqp (A14)






(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)eminusiP middotxv+

=

(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)ipart





Mpq = xppart

partxqminus xq

part

partxp(A16)


1

2MpqMqp(r

AY lm(x)) =

rA

2MpqMqpY


m(x) (A17)





m(x) (A19)

Thus we have

C2 rAY l




m(x) (A21)





F1 =1

λlog

((1 + ρs)


)(B1)

ndash 28 ndash

JHEP03(2016)165


of s to obtain

F1 =log(1+u)

u+

s


s2

u5(3u(2+u)minus 6 + 6u+ u2) log(1+u)

)+s3


)(B2)


F1 =

(minus1 +

u

2minus u2

3+u3


)+ s

(minus1

6+u

6minus 3u2


)+ s2

(minus1

30+

u

20minus 2u2


)+s3

(minus1

140+

u

70minus 5u2



The term at sk is

sk









in Pk(u) is

(2k)

(k)2(B6)


Qk(u) =ksuml=0

blul (B7)


ksuml=0

bl(minus1)l

(iminus l)= 0 (B8)




b1 =5b02 b2 =

20b09 b3 =

5b06 b4 =

5b042 b5 =

b0252

(B9)

ndash 29 ndash

JHEP03(2016)165



ksuml=0

MIlbl = 0 (B10)

where

MIl =(minus1)l




MIlbl (B12)



bl =sumI




bIb0 = 1 16bIb0 = 32 35 120bIb0 = 2 97 27 170

bIb0 =

5

2

20

95

6

5

42

1

252

bIb0 =

3

75

22

20

11

5

11

1

22

1

924

bIb0 =

7

2

63

13175

52

175

143

63

286

7

429

1

3432

(B14)



b1(k)b0 = k2

bk(k)b0 =k2

(2k)

bkminus1(k)b0 =k2

(2k)k(k + 1)

bkminus2(k)b0 =k2

(2k)

(k minus 1)k(k + 1)(k + 2)

4

bkminus3(k)b0 =k2

(2k)

(k

3

)(k + 3

3

)bkminusm(k)b0 =

k2

(2k)

(k

m

)(k +m

m

)(B15)

ndash 30 ndash

JHEP03(2016)165


so that

we have

b1(k) = k2(2k)

k2

bk(k) = 1



4

bkminus3(k) =

(k

3

)(k + 3

3

)bkminusm(k) =

(k

m

)(k +m

m

)(B16)


Pk(u) =

ksumi=1

aiuiminus1 (B17)



bl(minus1)iminusl

(iminus l)= 0 (B18)




k = 2 ~a = a1 ak = 6 3

k = 3 ~a =1

3times 60 60 11

k = 4 ~a =5

6times 84 126 52 5

k = 5 ~a =1

30times 7560 15120 9870 2310 137

k = 6 ~a =7

10times 13720 3300 2960 1140 174 7 (B19)

which leads to

a1(k) =(2k)

(k)2

ak(k) = 2h(k) = 2

ksuml=1

1

l(B20)




ndash 31 ndash

JHEP03(2016)165


grals






infinsumli=0

summi

infinsumt=0

sump


(x1)Yl2minusm2

(x2)Yl3m3

(xprime3)Yl4m4



t+ 1

=x23x

24

x213x224

1

λlog

((1 + ρs)


)(B21)



C l1 l2 lm1m2 m1+m2

=

langl12

l22

mL12

mL22

∣∣∣∣∣ l2

mL1 +mL2

2

ranglangl12

l22

mR12

mR22

∣∣∣∣∣ l2

mR1 +mR22






Yn12

n12n12

(x1)Yn22

n22n22



(x4) = zn11 zn2

2 zn33 zn4

4 (B22)


zn11 zn2

2 zn33 zn4


note that





that zn11 zn2

2 zn33 zn4


Yn12

n12n1minus2

2

(x1)Yn22

n2minus22

n22

(x2)Yn32


2

(x3)Yn42

minusn4minus22

minusn42

(x4)

=radicn1n2n3n4z

n1minus11 w1z

n2minus12 w2z

n3minus13 w3z

n4minus14 w4 (B24)


coefficients

Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B25)

Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B26)

ndash 32 ndash

JHEP03(2016)165


Cn12n22n1+n2

2n12n2minus2

2n1+n2minus2

2

=

radicn2

n1 + n2(B27)

Cn12n22n1+n2minus2

2n12n2minus2

2n1+n2minus2

2

= minusradic

n1n1 + n2

(B28)



n2minus12 w2z

n3minus13 w3z

n4minus14 w4 is

radicn1n2n3n4

[Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2 + 1

+Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2

]=


(B29)




section 5






parth








prime4vminus) (C1)



(2ixprime3microx

prime3 middot

part


part


)eiKmiddotx

prime3vminus

= ix23part

partxmicro3

1

x23eiKmiddotx

prime3vminus (C2)


parth


part

partxmicro2

(1

x22h

)minus x23

part

partxmicro3

(1

x23h

)minus x24

part

partxmicro4

(1

x24h

)(C3)

ndash 33 ndash

JHEP03(2016)165


x22x23x

24

(part

partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)[h

x22x23x

24

]= 0 (C4)


x1micropart

partx1νminus x1ν

part

partx1micro

)h= F1(Mmicroνe









x22x23x

24

(x1micro

part

partx1νminus x1ν

part


part

partx2νminus x2ν

part


part

partx3νminus x3ν

part

partx3micro

+x4micropart

partx4νminus x4ν

part

partx4micro

)[h

x22x23x

24

]= 0 (C6)


consider(x1 middot

part

partx1+ 3

)h = F1(e






x22x23x

24

(x1 middot

part

partx1+ x2 middot

part

partx2+ x3 middot

part

partx3+ x4 middot

part

partx4+ 12

)[h

x22x23x

24

]= 0 (C8)


h

x22x23x

24

=A




we have




ndash 34 ndash

JHEP03(2016)165





infinsumnm=0

im(minus1)n




where







minus (C14)

we find


and

infinsumn=0

(minus1)n




1


(C16)



=infinsum

n1n3n4=0

1



slot we get

α(P middot x2) =1


ndash 35 ndash

JHEP03(2016)165




=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)


=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)

(n1 + n3 + n4)



=infinsumn1

(minus1)n1




timesinfinsumn3

(minus1)n3




timesinfinsumn4

(minus1)n4





=1


1


1


=x22x

33x

24






References














ndash 36 ndash

JHEP03(2016)165










(2003) 013 [hep-th0212208] [INSPIRE]




(2014) 302 [arXiv14036646] [INSPIRE]






















(2015) 006 [arXiv150208005] [INSPIRE]






ndash 37 ndash

JHEP03(2016)165



[INSPIRE]

















(2015) 29FT01 [arXiv150500963] [INSPIRE]








[INSPIRE]














ndash 38 ndash

JHEP03(2016)165















ndash 39 ndash

Introduction























JHEP03(2016)165




I4 = I43 + I44 (319)


I5 =sumlimi

Cm1m2m3m4

l1l2l3l4

prodi

Y limi(xi)

int infin|x4|


=sumlimil

Cm1m2m3m4

l1l2l3l4

prodi

Y limi(xi)

(|x4|minus4minusl

4 + l

)δ(l l1 + l2 + l3 + l4) (320)






C2 = ∆(∆minus 4) + l(l + 2) = minus3 (321)










and Libine [29]


result is

log(r3r2

)sumlimi

(Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

)δ(l1 + l2 l3 + l4) (41)


Φ(s t) = F0(s t) + log(s)F1(s t) (42)



ndash 13 ndash

JHEP03(2016)165





)spart

parts

(spartH

parts

)minus(


(tpartH

partt

)+ 4s

partH

parts(44)


C2 sΦ = 2 sΦ C2 sF1 = 2 sF1 (45)



for the integral is

1

2x213x224

F1(s t) (46)



plies that

2sumlimi

(Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

)δ(l1 + l2 l3 + l4) =

1

x213x224

F1(s t)

(47)




V+ otimes V+ = A[200] +

infinoplusk=1

D[k+2 k2 k2] (48)








P++++ defined by




ndash 14 ndash

JHEP03(2016)165



















1

2x213x224



prime24 P(x1 x2 x










ndash 15 ndash

JHEP03(2016)165





m1(x)Y l2

m2(x)Y l5

m5(x) (416)


(x1)Yl2m2


Y l1m1

(x1)otimes Y l2m2


(x)Y l2m2

(x)

Y l3m3

(xprime3)

|x3|2otimesY l4m4

(xprime4)

|x4|2rarr

Y l3m3

(xprime)

|x|2otimesY l4m4

(xprime)

|x|2(417)















m(K)Y ml (xprime)

sumlimi

(minusi)l12l1 l1

Y l1m1


Y l2m2

(x2)(i)l3

2l3 l3Y l3m3

(xprime3)(i)l4

2l4 l4Y l4m4

(xprime4)

P++minusminus

(Y m1l1




(K)vminus)

=sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

P++minusminus

(Y m1l1

(P )

2l1 l1v+ otimes

Y m2l2

(P )

2l2 l2v+ otimes

Y m3l3

(K)

2l3 l3vminus otimes

Y m4l4

(K)

2l4 l4vminus)

(418)








ndash 16 ndash

JHEP03(2016)165






F1 =1

λlog

((1 + ρs)


)(419)



sk

u2k+1[Qk(u) log(1 + u)]+ (420)



Qk(u) =

ksuml=0

bl(k)ul (421)

where

bl(k) =

(k

k minus l

)(2k + l

k minus l

)(422)








equivariant maps


I = I1 + I2 + I3 + I4 + I5



I(1) = (IS11 + IS21)

I(2) = (IS22 + IS32 + IL32)

I(3) = (IL33 + IS33 + IS43)

I(4) = (IS45 + IS55) (52)

ndash 17 ndash

JHEP03(2016)165




I = I(1) + I(2) + I(3) + I(4) (53)



jI(i) = minus2δij

prodk 6=j

1

(xk minus xi)2(54)









I(1) =sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

r22

Y l3m3

(xprime3)

r23

Y l4m4

(xprime4)



(l1 + l2 + l3 + l4 + 4)(minusl1 + l2 + l3 + l4 + 2)(55)




part2


(r21)A = 4A(A+ 1)(r21)Aminus1

part


part

partxmicro1Y l1m1

(x1)part


21)Aminus1Y l1

m1(x1) (56)


part2


I(1) =sumlimi


r22r23r

24

Y l1m1

(x1)Yl2m2

(xprime2)Yl3m3

(xprime3)Yl4m4


(57)


1


1

r22r23r

24

sumlimi

Y l2m2

(x1)Yl3m3

(x1)Yl4m4

(x1)Yl2m2

(xprime2)Yl3m3

(xprime3)Yl4m4

(xprime4)

=(l1 + 1)rl2+l3+l41

r22r23r

24

sumlimi

Cm1m2m3m4l1l2l3l4

Y l1m1

(x)Y l2m2

(xprime2)Yl3m3

(xprime3)Yl4m4

(xprime4) (58)

ndash 18 ndash

JHEP03(2016)165














I(2) =sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


1

l2+l3+l4minusl1+2

δ(l1 + l2 l3 + l4)

r23r24

+sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


1

r23r24

times



]


part

partxmicro2Y l2m2

(x2)part


22)Aminus1Y l2

m2(x2)

part

partxmicro2logr2 =

xmicro2r22

part

partxmicro2

part

partxmicro2logr2 =

2

r22(59)

We find

part2




r23r24

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


(510)


minus 2







ndash 19 ndash

JHEP03(2016)165


equivariant map


such that



x22x23x

24








23x

24


prime2)f(x1 x

prime3)f(x1 x

prime4)

(515)



1


(516)







part2



prime2)f(x1 x

prime3)f(x1 x

prime4) (518)


24I

(1)(x1 x2 x3 x4)


x22x23x

24 I(1)(x1 x2 x3 x4)rarr F1(e


While x23x24I








ndash 20 ndash

JHEP03(2016)165

implies that (part

partxmicro1+ x22

part

partxmicro2

1

x22+ x23

part

partxmicro3

1

x23+ x24

part

partxmicro4

1

x24

)F1 = 0 (521)


24I



24(I

(1)+I(2)+I(3)+I(4)) = x22x23x



partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)I = 0 (522)





23x

24I(x1 x2 x3 x4) =

x22x23x

24

x213x224

Φ(s t)

(523)



with


minus)=xprime21 x

prime22 xprime23


The x21x22x

23I













〈φ(x1)φ3(x2)φ(x3)φ(x4)〉 (528)




ndash 21 ndash

JHEP03(2016)165



twisted co-product










23x

24



















prime22 xprime23






ndash 22 ndash

JHEP03(2016)165





I2 =

intd4x5d

4x6xminus256

x215x225x

245x

226x

246x

236

(61)






















ndash 23 ndash

JHEP03(2016)165
































ndash 24 ndash

JHEP03(2016)165































Acknowledgments








this work was done

ndash 25 ndash

JHEP03(2016)165








1

2π2

intdS3Y l1

m1(x)Y l2

m2(x) = gm1m2

δl1l2

(l1 + 1)(A1)




gm1m2 = δm1m2


m2(x) = Y l

m1(x)

Y ml (x) = Y l

m(x) (A2)






(mLmR)

gm1m2 = δm1minusm2

Y ml (x) = Y l

minusm(x) (A3)



ables are

z = x1 + ix2 harr(

1

2

1

2


(minus1

2minus1

2

)w = x3 + ix4 harr

(1

2minus1

2


(minus1

21

2

)(A4)

We find that

Y l(l2l2) = (x1 + ix2)

l = zl (A5)

has normalization

1

2π2

intdS3(Y l

( l2 l2))lowast Y l

( l2 l2)

=1

(l + 1)(A6)


ndash 26 ndash

JHEP03(2016)165




(minusi)l

2llY lm(x)Y m

l (P )v+ (A7)


η(Y l1m1

(P )v+ Y m2l2

(K)vminus)

= δl1l2δm1m222l1(l1)2 (A8)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)= δl1l2δm1m2 (A9)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)=

1

2π2

intS3

Y m2l2

(xprime)


m1(x)) (A10)





Y lm(xprime)

xprime2


m(x) (A12)


η = f η


1


summ

1

y2Y lm(x)Y m


=infinsuml=0

summ

1

x2Y lm(xprime)Y m

l (y) if |x| gt |y| (A13)






ndash 27 ndash

JHEP03(2016)165


C2 = minus1

2LMNL


2MpqMqp (A14)






(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)eminusiP middotxv+

=

(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)ipart





Mpq = xppart

partxqminus xq

part

partxp(A16)


1

2MpqMqp(r

AY lm(x)) =

rA

2MpqMqpY


m(x) (A17)





m(x) (A19)

Thus we have

C2 rAY l




m(x) (A21)





F1 =1

λlog

((1 + ρs)


)(B1)

ndash 28 ndash

JHEP03(2016)165


of s to obtain

F1 =log(1+u)

u+

s


s2

u5(3u(2+u)minus 6 + 6u+ u2) log(1+u)

)+s3


)(B2)


F1 =

(minus1 +

u

2minus u2

3+u3


)+ s

(minus1

6+u

6minus 3u2


)+ s2

(minus1

30+

u

20minus 2u2


)+s3

(minus1

140+

u

70minus 5u2



The term at sk is

sk









in Pk(u) is

(2k)

(k)2(B6)


Qk(u) =ksuml=0

blul (B7)


ksuml=0

bl(minus1)l

(iminus l)= 0 (B8)




b1 =5b02 b2 =

20b09 b3 =

5b06 b4 =

5b042 b5 =

b0252

(B9)

ndash 29 ndash

JHEP03(2016)165



ksuml=0

MIlbl = 0 (B10)

where

MIl =(minus1)l




MIlbl (B12)



bl =sumI




bIb0 = 1 16bIb0 = 32 35 120bIb0 = 2 97 27 170

bIb0 =

5

2

20

95

6

5

42

1

252

bIb0 =

3

75

22

20

11

5

11

1

22

1

924

bIb0 =

7

2

63

13175

52

175

143

63

286

7

429

1

3432

(B14)



b1(k)b0 = k2

bk(k)b0 =k2

(2k)

bkminus1(k)b0 =k2

(2k)k(k + 1)

bkminus2(k)b0 =k2

(2k)

(k minus 1)k(k + 1)(k + 2)

4

bkminus3(k)b0 =k2

(2k)

(k

3

)(k + 3

3

)bkminusm(k)b0 =

k2

(2k)

(k

m

)(k +m

m

)(B15)

ndash 30 ndash

JHEP03(2016)165


so that

we have

b1(k) = k2(2k)

k2

bk(k) = 1



4

bkminus3(k) =

(k

3

)(k + 3

3

)bkminusm(k) =

(k

m

)(k +m

m

)(B16)


Pk(u) =

ksumi=1

aiuiminus1 (B17)



bl(minus1)iminusl

(iminus l)= 0 (B18)




k = 2 ~a = a1 ak = 6 3

k = 3 ~a =1

3times 60 60 11

k = 4 ~a =5

6times 84 126 52 5

k = 5 ~a =1

30times 7560 15120 9870 2310 137

k = 6 ~a =7

10times 13720 3300 2960 1140 174 7 (B19)

which leads to

a1(k) =(2k)

(k)2

ak(k) = 2h(k) = 2

ksuml=1

1

l(B20)




ndash 31 ndash

JHEP03(2016)165


grals






infinsumli=0

summi

infinsumt=0

sump


(x1)Yl2minusm2

(x2)Yl3m3

(xprime3)Yl4m4



t+ 1

=x23x

24

x213x224

1

λlog

((1 + ρs)


)(B21)



C l1 l2 lm1m2 m1+m2

=

langl12

l22

mL12

mL22

∣∣∣∣∣ l2

mL1 +mL2

2

ranglangl12

l22

mR12

mR22

∣∣∣∣∣ l2

mR1 +mR22






Yn12

n12n12

(x1)Yn22

n22n22



(x4) = zn11 zn2

2 zn33 zn4

4 (B22)


zn11 zn2

2 zn33 zn4


note that





that zn11 zn2

2 zn33 zn4


Yn12

n12n1minus2

2

(x1)Yn22

n2minus22

n22

(x2)Yn32


2

(x3)Yn42

minusn4minus22

minusn42

(x4)

=radicn1n2n3n4z

n1minus11 w1z

n2minus12 w2z

n3minus13 w3z

n4minus14 w4 (B24)


coefficients

Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B25)

Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B26)

ndash 32 ndash

JHEP03(2016)165


Cn12n22n1+n2

2n12n2minus2

2n1+n2minus2

2

=

radicn2

n1 + n2(B27)

Cn12n22n1+n2minus2

2n12n2minus2

2n1+n2minus2

2

= minusradic

n1n1 + n2

(B28)



n2minus12 w2z

n3minus13 w3z

n4minus14 w4 is

radicn1n2n3n4

[Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2 + 1

+Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2

]=


(B29)




section 5






parth








prime4vminus) (C1)



(2ixprime3microx

prime3 middot

part


part


)eiKmiddotx

prime3vminus

= ix23part

partxmicro3

1

x23eiKmiddotx

prime3vminus (C2)


parth


part

partxmicro2

(1

x22h

)minus x23

part

partxmicro3

(1

x23h

)minus x24

part

partxmicro4

(1

x24h

)(C3)

ndash 33 ndash

JHEP03(2016)165


x22x23x

24

(part

partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)[h

x22x23x

24

]= 0 (C4)


x1micropart

partx1νminus x1ν

part

partx1micro

)h= F1(Mmicroνe









x22x23x

24

(x1micro

part

partx1νminus x1ν

part


part

partx2νminus x2ν

part


part

partx3νminus x3ν

part

partx3micro

+x4micropart

partx4νminus x4ν

part

partx4micro

)[h

x22x23x

24

]= 0 (C6)


consider(x1 middot

part

partx1+ 3

)h = F1(e






x22x23x

24

(x1 middot

part

partx1+ x2 middot

part

partx2+ x3 middot

part

partx3+ x4 middot

part

partx4+ 12

)[h

x22x23x

24

]= 0 (C8)


h

x22x23x

24

=A




we have




ndash 34 ndash

JHEP03(2016)165





infinsumnm=0

im(minus1)n




where







minus (C14)

we find


and

infinsumn=0

(minus1)n




1


(C16)



=infinsum

n1n3n4=0

1



slot we get

α(P middot x2) =1


ndash 35 ndash

JHEP03(2016)165




=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)


=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)

(n1 + n3 + n4)



=infinsumn1

(minus1)n1




timesinfinsumn3

(minus1)n3




timesinfinsumn4

(minus1)n4





=1


1


1


=x22x

33x

24






References














ndash 36 ndash

JHEP03(2016)165










(2003) 013 [hep-th0212208] [INSPIRE]




(2014) 302 [arXiv14036646] [INSPIRE]






















(2015) 006 [arXiv150208005] [INSPIRE]






ndash 37 ndash

JHEP03(2016)165



[INSPIRE]

















(2015) 29FT01 [arXiv150500963] [INSPIRE]








[INSPIRE]














ndash 38 ndash

JHEP03(2016)165















ndash 39 ndash

Introduction























JHEP03(2016)165





)spart

parts

(spartH

parts

)minus(


(tpartH

partt

)+ 4s

partH

parts(44)


C2 sΦ = 2 sΦ C2 sF1 = 2 sF1 (45)



for the integral is

1

2x213x224

F1(s t) (46)



plies that

2sumlimi

(Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

|x3|2|x4|2Cm1m2m3m4

l1l2l3l4

)δ(l1 + l2 l3 + l4) =

1

x213x224

F1(s t)

(47)




V+ otimes V+ = A[200] +

infinoplusk=1

D[k+2 k2 k2] (48)








P++++ defined by




ndash 14 ndash

JHEP03(2016)165



















1

2x213x224



prime24 P(x1 x2 x










ndash 15 ndash

JHEP03(2016)165





m1(x)Y l2

m2(x)Y l5

m5(x) (416)


(x1)Yl2m2


Y l1m1

(x1)otimes Y l2m2


(x)Y l2m2

(x)

Y l3m3

(xprime3)

|x3|2otimesY l4m4

(xprime4)

|x4|2rarr

Y l3m3

(xprime)

|x|2otimesY l4m4

(xprime)

|x|2(417)















m(K)Y ml (xprime)

sumlimi

(minusi)l12l1 l1

Y l1m1


Y l2m2

(x2)(i)l3

2l3 l3Y l3m3

(xprime3)(i)l4

2l4 l4Y l4m4

(xprime4)

P++minusminus

(Y m1l1




(K)vminus)

=sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

P++minusminus

(Y m1l1

(P )

2l1 l1v+ otimes

Y m2l2

(P )

2l2 l2v+ otimes

Y m3l3

(K)

2l3 l3vminus otimes

Y m4l4

(K)

2l4 l4vminus)

(418)








ndash 16 ndash

JHEP03(2016)165






F1 =1

λlog

((1 + ρs)


)(419)



sk

u2k+1[Qk(u) log(1 + u)]+ (420)



Qk(u) =

ksuml=0

bl(k)ul (421)

where

bl(k) =

(k

k minus l

)(2k + l

k minus l

)(422)








equivariant maps


I = I1 + I2 + I3 + I4 + I5



I(1) = (IS11 + IS21)

I(2) = (IS22 + IS32 + IL32)

I(3) = (IL33 + IS33 + IS43)

I(4) = (IS45 + IS55) (52)

ndash 17 ndash

JHEP03(2016)165




I = I(1) + I(2) + I(3) + I(4) (53)



jI(i) = minus2δij

prodk 6=j

1

(xk minus xi)2(54)









I(1) =sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

r22

Y l3m3

(xprime3)

r23

Y l4m4

(xprime4)



(l1 + l2 + l3 + l4 + 4)(minusl1 + l2 + l3 + l4 + 2)(55)




part2


(r21)A = 4A(A+ 1)(r21)Aminus1

part


part

partxmicro1Y l1m1

(x1)part


21)Aminus1Y l1

m1(x1) (56)


part2


I(1) =sumlimi


r22r23r

24

Y l1m1

(x1)Yl2m2

(xprime2)Yl3m3

(xprime3)Yl4m4


(57)


1


1

r22r23r

24

sumlimi

Y l2m2

(x1)Yl3m3

(x1)Yl4m4

(x1)Yl2m2

(xprime2)Yl3m3

(xprime3)Yl4m4

(xprime4)

=(l1 + 1)rl2+l3+l41

r22r23r

24

sumlimi

Cm1m2m3m4l1l2l3l4

Y l1m1

(x)Y l2m2

(xprime2)Yl3m3

(xprime3)Yl4m4

(xprime4) (58)

ndash 18 ndash

JHEP03(2016)165














I(2) =sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


1

l2+l3+l4minusl1+2

δ(l1 + l2 l3 + l4)

r23r24

+sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


1

r23r24

times



]


part

partxmicro2Y l2m2

(x2)part


22)Aminus1Y l2

m2(x2)

part

partxmicro2logr2 =

xmicro2r22

part

partxmicro2

part

partxmicro2logr2 =

2

r22(59)

We find

part2




r23r24

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


(510)


minus 2







ndash 19 ndash

JHEP03(2016)165


equivariant map


such that



x22x23x

24








23x

24


prime2)f(x1 x

prime3)f(x1 x

prime4)

(515)



1


(516)







part2



prime2)f(x1 x

prime3)f(x1 x

prime4) (518)


24I

(1)(x1 x2 x3 x4)


x22x23x

24 I(1)(x1 x2 x3 x4)rarr F1(e


While x23x24I








ndash 20 ndash

JHEP03(2016)165

implies that (part

partxmicro1+ x22

part

partxmicro2

1

x22+ x23

part

partxmicro3

1

x23+ x24

part

partxmicro4

1

x24

)F1 = 0 (521)


24I



24(I

(1)+I(2)+I(3)+I(4)) = x22x23x



partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)I = 0 (522)





23x

24I(x1 x2 x3 x4) =

x22x23x

24

x213x224

Φ(s t)

(523)



with


minus)=xprime21 x

prime22 xprime23


The x21x22x

23I













〈φ(x1)φ3(x2)φ(x3)φ(x4)〉 (528)




ndash 21 ndash

JHEP03(2016)165



twisted co-product










23x

24



















prime22 xprime23






ndash 22 ndash

JHEP03(2016)165





I2 =

intd4x5d

4x6xminus256

x215x225x

245x

226x

246x

236

(61)






















ndash 23 ndash

JHEP03(2016)165
































ndash 24 ndash

JHEP03(2016)165































Acknowledgments








this work was done

ndash 25 ndash

JHEP03(2016)165








1

2π2

intdS3Y l1

m1(x)Y l2

m2(x) = gm1m2

δl1l2

(l1 + 1)(A1)




gm1m2 = δm1m2


m2(x) = Y l

m1(x)

Y ml (x) = Y l

m(x) (A2)






(mLmR)

gm1m2 = δm1minusm2

Y ml (x) = Y l

minusm(x) (A3)



ables are

z = x1 + ix2 harr(

1

2

1

2


(minus1

2minus1

2

)w = x3 + ix4 harr

(1

2minus1

2


(minus1

21

2

)(A4)

We find that

Y l(l2l2) = (x1 + ix2)

l = zl (A5)

has normalization

1

2π2

intdS3(Y l

( l2 l2))lowast Y l

( l2 l2)

=1

(l + 1)(A6)


ndash 26 ndash

JHEP03(2016)165




(minusi)l

2llY lm(x)Y m

l (P )v+ (A7)


η(Y l1m1

(P )v+ Y m2l2

(K)vminus)

= δl1l2δm1m222l1(l1)2 (A8)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)= δl1l2δm1m2 (A9)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)=

1

2π2

intS3

Y m2l2

(xprime)


m1(x)) (A10)





Y lm(xprime)

xprime2


m(x) (A12)


η = f η


1


summ

1

y2Y lm(x)Y m


=infinsuml=0

summ

1

x2Y lm(xprime)Y m

l (y) if |x| gt |y| (A13)






ndash 27 ndash

JHEP03(2016)165


C2 = minus1

2LMNL


2MpqMqp (A14)






(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)eminusiP middotxv+

=

(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)ipart





Mpq = xppart

partxqminus xq

part

partxp(A16)


1

2MpqMqp(r

AY lm(x)) =

rA

2MpqMqpY


m(x) (A17)





m(x) (A19)

Thus we have

C2 rAY l




m(x) (A21)





F1 =1

λlog

((1 + ρs)


)(B1)

ndash 28 ndash

JHEP03(2016)165


of s to obtain

F1 =log(1+u)

u+

s


s2

u5(3u(2+u)minus 6 + 6u+ u2) log(1+u)

)+s3


)(B2)


F1 =

(minus1 +

u

2minus u2

3+u3


)+ s

(minus1

6+u

6minus 3u2


)+ s2

(minus1

30+

u

20minus 2u2


)+s3

(minus1

140+

u

70minus 5u2



The term at sk is

sk









in Pk(u) is

(2k)

(k)2(B6)


Qk(u) =ksuml=0

blul (B7)


ksuml=0

bl(minus1)l

(iminus l)= 0 (B8)




b1 =5b02 b2 =

20b09 b3 =

5b06 b4 =

5b042 b5 =

b0252

(B9)

ndash 29 ndash

JHEP03(2016)165



ksuml=0

MIlbl = 0 (B10)

where

MIl =(minus1)l




MIlbl (B12)



bl =sumI




bIb0 = 1 16bIb0 = 32 35 120bIb0 = 2 97 27 170

bIb0 =

5

2

20

95

6

5

42

1

252

bIb0 =

3

75

22

20

11

5

11

1

22

1

924

bIb0 =

7

2

63

13175

52

175

143

63

286

7

429

1

3432

(B14)



b1(k)b0 = k2

bk(k)b0 =k2

(2k)

bkminus1(k)b0 =k2

(2k)k(k + 1)

bkminus2(k)b0 =k2

(2k)

(k minus 1)k(k + 1)(k + 2)

4

bkminus3(k)b0 =k2

(2k)

(k

3

)(k + 3

3

)bkminusm(k)b0 =

k2

(2k)

(k

m

)(k +m

m

)(B15)

ndash 30 ndash

JHEP03(2016)165


so that

we have

b1(k) = k2(2k)

k2

bk(k) = 1



4

bkminus3(k) =

(k

3

)(k + 3

3

)bkminusm(k) =

(k

m

)(k +m

m

)(B16)


Pk(u) =

ksumi=1

aiuiminus1 (B17)



bl(minus1)iminusl

(iminus l)= 0 (B18)




k = 2 ~a = a1 ak = 6 3

k = 3 ~a =1

3times 60 60 11

k = 4 ~a =5

6times 84 126 52 5

k = 5 ~a =1

30times 7560 15120 9870 2310 137

k = 6 ~a =7

10times 13720 3300 2960 1140 174 7 (B19)

which leads to

a1(k) =(2k)

(k)2

ak(k) = 2h(k) = 2

ksuml=1

1

l(B20)




ndash 31 ndash

JHEP03(2016)165


grals






infinsumli=0

summi

infinsumt=0

sump


(x1)Yl2minusm2

(x2)Yl3m3

(xprime3)Yl4m4



t+ 1

=x23x

24

x213x224

1

λlog

((1 + ρs)


)(B21)



C l1 l2 lm1m2 m1+m2

=

langl12

l22

mL12

mL22

∣∣∣∣∣ l2

mL1 +mL2

2

ranglangl12

l22

mR12

mR22

∣∣∣∣∣ l2

mR1 +mR22






Yn12

n12n12

(x1)Yn22

n22n22



(x4) = zn11 zn2

2 zn33 zn4

4 (B22)


zn11 zn2

2 zn33 zn4


note that





that zn11 zn2

2 zn33 zn4


Yn12

n12n1minus2

2

(x1)Yn22

n2minus22

n22

(x2)Yn32


2

(x3)Yn42

minusn4minus22

minusn42

(x4)

=radicn1n2n3n4z

n1minus11 w1z

n2minus12 w2z

n3minus13 w3z

n4minus14 w4 (B24)


coefficients

Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B25)

Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B26)

ndash 32 ndash

JHEP03(2016)165


Cn12n22n1+n2

2n12n2minus2

2n1+n2minus2

2

=

radicn2

n1 + n2(B27)

Cn12n22n1+n2minus2

2n12n2minus2

2n1+n2minus2

2

= minusradic

n1n1 + n2

(B28)



n2minus12 w2z

n3minus13 w3z

n4minus14 w4 is

radicn1n2n3n4

[Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2 + 1

+Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2

]=


(B29)




section 5






parth








prime4vminus) (C1)



(2ixprime3microx

prime3 middot

part


part


)eiKmiddotx

prime3vminus

= ix23part

partxmicro3

1

x23eiKmiddotx

prime3vminus (C2)


parth


part

partxmicro2

(1

x22h

)minus x23

part

partxmicro3

(1

x23h

)minus x24

part

partxmicro4

(1

x24h

)(C3)

ndash 33 ndash

JHEP03(2016)165


x22x23x

24

(part

partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)[h

x22x23x

24

]= 0 (C4)


x1micropart

partx1νminus x1ν

part

partx1micro

)h= F1(Mmicroνe









x22x23x

24

(x1micro

part

partx1νminus x1ν

part


part

partx2νminus x2ν

part


part

partx3νminus x3ν

part

partx3micro

+x4micropart

partx4νminus x4ν

part

partx4micro

)[h

x22x23x

24

]= 0 (C6)


consider(x1 middot

part

partx1+ 3

)h = F1(e






x22x23x

24

(x1 middot

part

partx1+ x2 middot

part

partx2+ x3 middot

part

partx3+ x4 middot

part

partx4+ 12

)[h

x22x23x

24

]= 0 (C8)


h

x22x23x

24

=A




we have




ndash 34 ndash

JHEP03(2016)165





infinsumnm=0

im(minus1)n




where







minus (C14)

we find


and

infinsumn=0

(minus1)n




1


(C16)



=infinsum

n1n3n4=0

1



slot we get

α(P middot x2) =1


ndash 35 ndash

JHEP03(2016)165




=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)


=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)

(n1 + n3 + n4)



=infinsumn1

(minus1)n1




timesinfinsumn3

(minus1)n3




timesinfinsumn4

(minus1)n4





=1


1


1


=x22x

33x

24






References














ndash 36 ndash

JHEP03(2016)165










(2003) 013 [hep-th0212208] [INSPIRE]




(2014) 302 [arXiv14036646] [INSPIRE]






















(2015) 006 [arXiv150208005] [INSPIRE]






ndash 37 ndash

JHEP03(2016)165



[INSPIRE]

















(2015) 29FT01 [arXiv150500963] [INSPIRE]








[INSPIRE]














ndash 38 ndash

JHEP03(2016)165















ndash 39 ndash

Introduction























JHEP03(2016)165



















1

2x213x224



prime24 P(x1 x2 x










ndash 15 ndash

JHEP03(2016)165





m1(x)Y l2

m2(x)Y l5

m5(x) (416)


(x1)Yl2m2


Y l1m1

(x1)otimes Y l2m2


(x)Y l2m2

(x)

Y l3m3

(xprime3)

|x3|2otimesY l4m4

(xprime4)

|x4|2rarr

Y l3m3

(xprime)

|x|2otimesY l4m4

(xprime)

|x|2(417)















m(K)Y ml (xprime)

sumlimi

(minusi)l12l1 l1

Y l1m1


Y l2m2

(x2)(i)l3

2l3 l3Y l3m3

(xprime3)(i)l4

2l4 l4Y l4m4

(xprime4)

P++minusminus

(Y m1l1




(K)vminus)

=sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

P++minusminus

(Y m1l1

(P )

2l1 l1v+ otimes

Y m2l2

(P )

2l2 l2v+ otimes

Y m3l3

(K)

2l3 l3vminus otimes

Y m4l4

(K)

2l4 l4vminus)

(418)








ndash 16 ndash

JHEP03(2016)165






F1 =1

λlog

((1 + ρs)


)(419)



sk

u2k+1[Qk(u) log(1 + u)]+ (420)



Qk(u) =

ksuml=0

bl(k)ul (421)

where

bl(k) =

(k

k minus l

)(2k + l

k minus l

)(422)








equivariant maps


I = I1 + I2 + I3 + I4 + I5



I(1) = (IS11 + IS21)

I(2) = (IS22 + IS32 + IL32)

I(3) = (IL33 + IS33 + IS43)

I(4) = (IS45 + IS55) (52)

ndash 17 ndash

JHEP03(2016)165




I = I(1) + I(2) + I(3) + I(4) (53)



jI(i) = minus2δij

prodk 6=j

1

(xk minus xi)2(54)









I(1) =sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

r22

Y l3m3

(xprime3)

r23

Y l4m4

(xprime4)



(l1 + l2 + l3 + l4 + 4)(minusl1 + l2 + l3 + l4 + 2)(55)




part2


(r21)A = 4A(A+ 1)(r21)Aminus1

part


part

partxmicro1Y l1m1

(x1)part


21)Aminus1Y l1

m1(x1) (56)


part2


I(1) =sumlimi


r22r23r

24

Y l1m1

(x1)Yl2m2

(xprime2)Yl3m3

(xprime3)Yl4m4


(57)


1


1

r22r23r

24

sumlimi

Y l2m2

(x1)Yl3m3

(x1)Yl4m4

(x1)Yl2m2

(xprime2)Yl3m3

(xprime3)Yl4m4

(xprime4)

=(l1 + 1)rl2+l3+l41

r22r23r

24

sumlimi

Cm1m2m3m4l1l2l3l4

Y l1m1

(x)Y l2m2

(xprime2)Yl3m3

(xprime3)Yl4m4

(xprime4) (58)

ndash 18 ndash

JHEP03(2016)165














I(2) =sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


1

l2+l3+l4minusl1+2

δ(l1 + l2 l3 + l4)

r23r24

+sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


1

r23r24

times



]


part

partxmicro2Y l2m2

(x2)part


22)Aminus1Y l2

m2(x2)

part

partxmicro2logr2 =

xmicro2r22

part

partxmicro2

part

partxmicro2logr2 =

2

r22(59)

We find

part2




r23r24

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


(510)


minus 2







ndash 19 ndash

JHEP03(2016)165


equivariant map


such that



x22x23x

24








23x

24


prime2)f(x1 x

prime3)f(x1 x

prime4)

(515)



1


(516)







part2



prime2)f(x1 x

prime3)f(x1 x

prime4) (518)


24I

(1)(x1 x2 x3 x4)


x22x23x

24 I(1)(x1 x2 x3 x4)rarr F1(e


While x23x24I








ndash 20 ndash

JHEP03(2016)165

implies that (part

partxmicro1+ x22

part

partxmicro2

1

x22+ x23

part

partxmicro3

1

x23+ x24

part

partxmicro4

1

x24

)F1 = 0 (521)


24I



24(I

(1)+I(2)+I(3)+I(4)) = x22x23x



partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)I = 0 (522)





23x

24I(x1 x2 x3 x4) =

x22x23x

24

x213x224

Φ(s t)

(523)



with


minus)=xprime21 x

prime22 xprime23


The x21x22x

23I













〈φ(x1)φ3(x2)φ(x3)φ(x4)〉 (528)




ndash 21 ndash

JHEP03(2016)165



twisted co-product










23x

24



















prime22 xprime23






ndash 22 ndash

JHEP03(2016)165





I2 =

intd4x5d

4x6xminus256

x215x225x

245x

226x

246x

236

(61)






















ndash 23 ndash

JHEP03(2016)165
































ndash 24 ndash

JHEP03(2016)165































Acknowledgments








this work was done

ndash 25 ndash

JHEP03(2016)165








1

2π2

intdS3Y l1

m1(x)Y l2

m2(x) = gm1m2

δl1l2

(l1 + 1)(A1)




gm1m2 = δm1m2


m2(x) = Y l

m1(x)

Y ml (x) = Y l

m(x) (A2)






(mLmR)

gm1m2 = δm1minusm2

Y ml (x) = Y l

minusm(x) (A3)



ables are

z = x1 + ix2 harr(

1

2

1

2


(minus1

2minus1

2

)w = x3 + ix4 harr

(1

2minus1

2


(minus1

21

2

)(A4)

We find that

Y l(l2l2) = (x1 + ix2)

l = zl (A5)

has normalization

1

2π2

intdS3(Y l

( l2 l2))lowast Y l

( l2 l2)

=1

(l + 1)(A6)


ndash 26 ndash

JHEP03(2016)165




(minusi)l

2llY lm(x)Y m

l (P )v+ (A7)


η(Y l1m1

(P )v+ Y m2l2

(K)vminus)

= δl1l2δm1m222l1(l1)2 (A8)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)= δl1l2δm1m2 (A9)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)=

1

2π2

intS3

Y m2l2

(xprime)


m1(x)) (A10)





Y lm(xprime)

xprime2


m(x) (A12)


η = f η


1


summ

1

y2Y lm(x)Y m


=infinsuml=0

summ

1

x2Y lm(xprime)Y m

l (y) if |x| gt |y| (A13)






ndash 27 ndash

JHEP03(2016)165


C2 = minus1

2LMNL


2MpqMqp (A14)






(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)eminusiP middotxv+

=

(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)ipart





Mpq = xppart

partxqminus xq

part

partxp(A16)


1

2MpqMqp(r

AY lm(x)) =

rA

2MpqMqpY


m(x) (A17)





m(x) (A19)

Thus we have

C2 rAY l




m(x) (A21)





F1 =1

λlog

((1 + ρs)


)(B1)

ndash 28 ndash

JHEP03(2016)165


of s to obtain

F1 =log(1+u)

u+

s


s2

u5(3u(2+u)minus 6 + 6u+ u2) log(1+u)

)+s3


)(B2)


F1 =

(minus1 +

u

2minus u2

3+u3


)+ s

(minus1

6+u

6minus 3u2


)+ s2

(minus1

30+

u

20minus 2u2


)+s3

(minus1

140+

u

70minus 5u2



The term at sk is

sk









in Pk(u) is

(2k)

(k)2(B6)


Qk(u) =ksuml=0

blul (B7)


ksuml=0

bl(minus1)l

(iminus l)= 0 (B8)




b1 =5b02 b2 =

20b09 b3 =

5b06 b4 =

5b042 b5 =

b0252

(B9)

ndash 29 ndash

JHEP03(2016)165



ksuml=0

MIlbl = 0 (B10)

where

MIl =(minus1)l




MIlbl (B12)



bl =sumI




bIb0 = 1 16bIb0 = 32 35 120bIb0 = 2 97 27 170

bIb0 =

5

2

20

95

6

5

42

1

252

bIb0 =

3

75

22

20

11

5

11

1

22

1

924

bIb0 =

7

2

63

13175

52

175

143

63

286

7

429

1

3432

(B14)



b1(k)b0 = k2

bk(k)b0 =k2

(2k)

bkminus1(k)b0 =k2

(2k)k(k + 1)

bkminus2(k)b0 =k2

(2k)

(k minus 1)k(k + 1)(k + 2)

4

bkminus3(k)b0 =k2

(2k)

(k

3

)(k + 3

3

)bkminusm(k)b0 =

k2

(2k)

(k

m

)(k +m

m

)(B15)

ndash 30 ndash

JHEP03(2016)165


so that

we have

b1(k) = k2(2k)

k2

bk(k) = 1



4

bkminus3(k) =

(k

3

)(k + 3

3

)bkminusm(k) =

(k

m

)(k +m

m

)(B16)


Pk(u) =

ksumi=1

aiuiminus1 (B17)



bl(minus1)iminusl

(iminus l)= 0 (B18)




k = 2 ~a = a1 ak = 6 3

k = 3 ~a =1

3times 60 60 11

k = 4 ~a =5

6times 84 126 52 5

k = 5 ~a =1

30times 7560 15120 9870 2310 137

k = 6 ~a =7

10times 13720 3300 2960 1140 174 7 (B19)

which leads to

a1(k) =(2k)

(k)2

ak(k) = 2h(k) = 2

ksuml=1

1

l(B20)




ndash 31 ndash

JHEP03(2016)165


grals






infinsumli=0

summi

infinsumt=0

sump


(x1)Yl2minusm2

(x2)Yl3m3

(xprime3)Yl4m4



t+ 1

=x23x

24

x213x224

1

λlog

((1 + ρs)


)(B21)



C l1 l2 lm1m2 m1+m2

=

langl12

l22

mL12

mL22

∣∣∣∣∣ l2

mL1 +mL2

2

ranglangl12

l22

mR12

mR22

∣∣∣∣∣ l2

mR1 +mR22






Yn12

n12n12

(x1)Yn22

n22n22



(x4) = zn11 zn2

2 zn33 zn4

4 (B22)


zn11 zn2

2 zn33 zn4


note that





that zn11 zn2

2 zn33 zn4


Yn12

n12n1minus2

2

(x1)Yn22

n2minus22

n22

(x2)Yn32


2

(x3)Yn42

minusn4minus22

minusn42

(x4)

=radicn1n2n3n4z

n1minus11 w1z

n2minus12 w2z

n3minus13 w3z

n4minus14 w4 (B24)


coefficients

Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B25)

Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B26)

ndash 32 ndash

JHEP03(2016)165


Cn12n22n1+n2

2n12n2minus2

2n1+n2minus2

2

=

radicn2

n1 + n2(B27)

Cn12n22n1+n2minus2

2n12n2minus2

2n1+n2minus2

2

= minusradic

n1n1 + n2

(B28)



n2minus12 w2z

n3minus13 w3z

n4minus14 w4 is

radicn1n2n3n4

[Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2 + 1

+Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2

]=


(B29)




section 5






parth








prime4vminus) (C1)



(2ixprime3microx

prime3 middot

part


part


)eiKmiddotx

prime3vminus

= ix23part

partxmicro3

1

x23eiKmiddotx

prime3vminus (C2)


parth


part

partxmicro2

(1

x22h

)minus x23

part

partxmicro3

(1

x23h

)minus x24

part

partxmicro4

(1

x24h

)(C3)

ndash 33 ndash

JHEP03(2016)165


x22x23x

24

(part

partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)[h

x22x23x

24

]= 0 (C4)


x1micropart

partx1νminus x1ν

part

partx1micro

)h= F1(Mmicroνe









x22x23x

24

(x1micro

part

partx1νminus x1ν

part


part

partx2νminus x2ν

part


part

partx3νminus x3ν

part

partx3micro

+x4micropart

partx4νminus x4ν

part

partx4micro

)[h

x22x23x

24

]= 0 (C6)


consider(x1 middot

part

partx1+ 3

)h = F1(e






x22x23x

24

(x1 middot

part

partx1+ x2 middot

part

partx2+ x3 middot

part

partx3+ x4 middot

part

partx4+ 12

)[h

x22x23x

24

]= 0 (C8)


h

x22x23x

24

=A




we have




ndash 34 ndash

JHEP03(2016)165





infinsumnm=0

im(minus1)n




where







minus (C14)

we find


and

infinsumn=0

(minus1)n




1


(C16)



=infinsum

n1n3n4=0

1



slot we get

α(P middot x2) =1


ndash 35 ndash

JHEP03(2016)165




=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)


=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)

(n1 + n3 + n4)



=infinsumn1

(minus1)n1




timesinfinsumn3

(minus1)n3




timesinfinsumn4

(minus1)n4





=1


1


1


=x22x

33x

24






References














ndash 36 ndash

JHEP03(2016)165










(2003) 013 [hep-th0212208] [INSPIRE]




(2014) 302 [arXiv14036646] [INSPIRE]






















(2015) 006 [arXiv150208005] [INSPIRE]






ndash 37 ndash

JHEP03(2016)165



[INSPIRE]

















(2015) 29FT01 [arXiv150500963] [INSPIRE]








[INSPIRE]














ndash 38 ndash

JHEP03(2016)165















ndash 39 ndash

Introduction























JHEP03(2016)165





m1(x)Y l2

m2(x)Y l5

m5(x) (416)


(x1)Yl2m2


Y l1m1

(x1)otimes Y l2m2


(x)Y l2m2

(x)

Y l3m3

(xprime3)

|x3|2otimesY l4m4

(xprime4)

|x4|2rarr

Y l3m3

(xprime)

|x|2otimesY l4m4

(xprime)

|x|2(417)















m(K)Y ml (xprime)

sumlimi

(minusi)l12l1 l1

Y l1m1


Y l2m2

(x2)(i)l3

2l3 l3Y l3m3

(xprime3)(i)l4

2l4 l4Y l4m4

(xprime4)

P++minusminus

(Y m1l1




(K)vminus)

=sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4

(xprime4)

P++minusminus

(Y m1l1

(P )

2l1 l1v+ otimes

Y m2l2

(P )

2l2 l2v+ otimes

Y m3l3

(K)

2l3 l3vminus otimes

Y m4l4

(K)

2l4 l4vminus)

(418)








ndash 16 ndash

JHEP03(2016)165






F1 =1

λlog

((1 + ρs)


)(419)



sk

u2k+1[Qk(u) log(1 + u)]+ (420)



Qk(u) =

ksuml=0

bl(k)ul (421)

where

bl(k) =

(k

k minus l

)(2k + l

k minus l

)(422)








equivariant maps


I = I1 + I2 + I3 + I4 + I5



I(1) = (IS11 + IS21)

I(2) = (IS22 + IS32 + IL32)

I(3) = (IL33 + IS33 + IS43)

I(4) = (IS45 + IS55) (52)

ndash 17 ndash

JHEP03(2016)165




I = I(1) + I(2) + I(3) + I(4) (53)



jI(i) = minus2δij

prodk 6=j

1

(xk minus xi)2(54)









I(1) =sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

r22

Y l3m3

(xprime3)

r23

Y l4m4

(xprime4)



(l1 + l2 + l3 + l4 + 4)(minusl1 + l2 + l3 + l4 + 2)(55)




part2


(r21)A = 4A(A+ 1)(r21)Aminus1

part


part

partxmicro1Y l1m1

(x1)part


21)Aminus1Y l1

m1(x1) (56)


part2


I(1) =sumlimi


r22r23r

24

Y l1m1

(x1)Yl2m2

(xprime2)Yl3m3

(xprime3)Yl4m4


(57)


1


1

r22r23r

24

sumlimi

Y l2m2

(x1)Yl3m3

(x1)Yl4m4

(x1)Yl2m2

(xprime2)Yl3m3

(xprime3)Yl4m4

(xprime4)

=(l1 + 1)rl2+l3+l41

r22r23r

24

sumlimi

Cm1m2m3m4l1l2l3l4

Y l1m1

(x)Y l2m2

(xprime2)Yl3m3

(xprime3)Yl4m4

(xprime4) (58)

ndash 18 ndash

JHEP03(2016)165














I(2) =sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


1

l2+l3+l4minusl1+2

δ(l1 + l2 l3 + l4)

r23r24

+sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


1

r23r24

times



]


part

partxmicro2Y l2m2

(x2)part


22)Aminus1Y l2

m2(x2)

part

partxmicro2logr2 =

xmicro2r22

part

partxmicro2

part

partxmicro2logr2 =

2

r22(59)

We find

part2




r23r24

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


(510)


minus 2







ndash 19 ndash

JHEP03(2016)165


equivariant map


such that



x22x23x

24








23x

24


prime2)f(x1 x

prime3)f(x1 x

prime4)

(515)



1


(516)







part2



prime2)f(x1 x

prime3)f(x1 x

prime4) (518)


24I

(1)(x1 x2 x3 x4)


x22x23x

24 I(1)(x1 x2 x3 x4)rarr F1(e


While x23x24I








ndash 20 ndash

JHEP03(2016)165

implies that (part

partxmicro1+ x22

part

partxmicro2

1

x22+ x23

part

partxmicro3

1

x23+ x24

part

partxmicro4

1

x24

)F1 = 0 (521)


24I



24(I

(1)+I(2)+I(3)+I(4)) = x22x23x



partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)I = 0 (522)





23x

24I(x1 x2 x3 x4) =

x22x23x

24

x213x224

Φ(s t)

(523)



with


minus)=xprime21 x

prime22 xprime23


The x21x22x

23I













〈φ(x1)φ3(x2)φ(x3)φ(x4)〉 (528)




ndash 21 ndash

JHEP03(2016)165



twisted co-product










23x

24



















prime22 xprime23






ndash 22 ndash

JHEP03(2016)165





I2 =

intd4x5d

4x6xminus256

x215x225x

245x

226x

246x

236

(61)






















ndash 23 ndash

JHEP03(2016)165
































ndash 24 ndash

JHEP03(2016)165































Acknowledgments








this work was done

ndash 25 ndash

JHEP03(2016)165








1

2π2

intdS3Y l1

m1(x)Y l2

m2(x) = gm1m2

δl1l2

(l1 + 1)(A1)




gm1m2 = δm1m2


m2(x) = Y l

m1(x)

Y ml (x) = Y l

m(x) (A2)






(mLmR)

gm1m2 = δm1minusm2

Y ml (x) = Y l

minusm(x) (A3)



ables are

z = x1 + ix2 harr(

1

2

1

2


(minus1

2minus1

2

)w = x3 + ix4 harr

(1

2minus1

2


(minus1

21

2

)(A4)

We find that

Y l(l2l2) = (x1 + ix2)

l = zl (A5)

has normalization

1

2π2

intdS3(Y l

( l2 l2))lowast Y l

( l2 l2)

=1

(l + 1)(A6)


ndash 26 ndash

JHEP03(2016)165




(minusi)l

2llY lm(x)Y m

l (P )v+ (A7)


η(Y l1m1

(P )v+ Y m2l2

(K)vminus)

= δl1l2δm1m222l1(l1)2 (A8)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)= δl1l2δm1m2 (A9)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)=

1

2π2

intS3

Y m2l2

(xprime)


m1(x)) (A10)





Y lm(xprime)

xprime2


m(x) (A12)


η = f η


1


summ

1

y2Y lm(x)Y m


=infinsuml=0

summ

1

x2Y lm(xprime)Y m

l (y) if |x| gt |y| (A13)






ndash 27 ndash

JHEP03(2016)165


C2 = minus1

2LMNL


2MpqMqp (A14)






(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)eminusiP middotxv+

=

(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)ipart





Mpq = xppart

partxqminus xq

part

partxp(A16)


1

2MpqMqp(r

AY lm(x)) =

rA

2MpqMqpY


m(x) (A17)





m(x) (A19)

Thus we have

C2 rAY l




m(x) (A21)





F1 =1

λlog

((1 + ρs)


)(B1)

ndash 28 ndash

JHEP03(2016)165


of s to obtain

F1 =log(1+u)

u+

s


s2

u5(3u(2+u)minus 6 + 6u+ u2) log(1+u)

)+s3


)(B2)


F1 =

(minus1 +

u

2minus u2

3+u3


)+ s

(minus1

6+u

6minus 3u2


)+ s2

(minus1

30+

u

20minus 2u2


)+s3

(minus1

140+

u

70minus 5u2



The term at sk is

sk









in Pk(u) is

(2k)

(k)2(B6)


Qk(u) =ksuml=0

blul (B7)


ksuml=0

bl(minus1)l

(iminus l)= 0 (B8)




b1 =5b02 b2 =

20b09 b3 =

5b06 b4 =

5b042 b5 =

b0252

(B9)

ndash 29 ndash

JHEP03(2016)165



ksuml=0

MIlbl = 0 (B10)

where

MIl =(minus1)l




MIlbl (B12)



bl =sumI




bIb0 = 1 16bIb0 = 32 35 120bIb0 = 2 97 27 170

bIb0 =

5

2

20

95

6

5

42

1

252

bIb0 =

3

75

22

20

11

5

11

1

22

1

924

bIb0 =

7

2

63

13175

52

175

143

63

286

7

429

1

3432

(B14)



b1(k)b0 = k2

bk(k)b0 =k2

(2k)

bkminus1(k)b0 =k2

(2k)k(k + 1)

bkminus2(k)b0 =k2

(2k)

(k minus 1)k(k + 1)(k + 2)

4

bkminus3(k)b0 =k2

(2k)

(k

3

)(k + 3

3

)bkminusm(k)b0 =

k2

(2k)

(k

m

)(k +m

m

)(B15)

ndash 30 ndash

JHEP03(2016)165


so that

we have

b1(k) = k2(2k)

k2

bk(k) = 1



4

bkminus3(k) =

(k

3

)(k + 3

3

)bkminusm(k) =

(k

m

)(k +m

m

)(B16)


Pk(u) =

ksumi=1

aiuiminus1 (B17)



bl(minus1)iminusl

(iminus l)= 0 (B18)




k = 2 ~a = a1 ak = 6 3

k = 3 ~a =1

3times 60 60 11

k = 4 ~a =5

6times 84 126 52 5

k = 5 ~a =1

30times 7560 15120 9870 2310 137

k = 6 ~a =7

10times 13720 3300 2960 1140 174 7 (B19)

which leads to

a1(k) =(2k)

(k)2

ak(k) = 2h(k) = 2

ksuml=1

1

l(B20)




ndash 31 ndash

JHEP03(2016)165


grals






infinsumli=0

summi

infinsumt=0

sump


(x1)Yl2minusm2

(x2)Yl3m3

(xprime3)Yl4m4



t+ 1

=x23x

24

x213x224

1

λlog

((1 + ρs)


)(B21)



C l1 l2 lm1m2 m1+m2

=

langl12

l22

mL12

mL22

∣∣∣∣∣ l2

mL1 +mL2

2

ranglangl12

l22

mR12

mR22

∣∣∣∣∣ l2

mR1 +mR22






Yn12

n12n12

(x1)Yn22

n22n22



(x4) = zn11 zn2

2 zn33 zn4

4 (B22)


zn11 zn2

2 zn33 zn4


note that





that zn11 zn2

2 zn33 zn4


Yn12

n12n1minus2

2

(x1)Yn22

n2minus22

n22

(x2)Yn32


2

(x3)Yn42

minusn4minus22

minusn42

(x4)

=radicn1n2n3n4z

n1minus11 w1z

n2minus12 w2z

n3minus13 w3z

n4minus14 w4 (B24)


coefficients

Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B25)

Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B26)

ndash 32 ndash

JHEP03(2016)165


Cn12n22n1+n2

2n12n2minus2

2n1+n2minus2

2

=

radicn2

n1 + n2(B27)

Cn12n22n1+n2minus2

2n12n2minus2

2n1+n2minus2

2

= minusradic

n1n1 + n2

(B28)



n2minus12 w2z

n3minus13 w3z

n4minus14 w4 is

radicn1n2n3n4

[Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2 + 1

+Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2

]=


(B29)




section 5






parth








prime4vminus) (C1)



(2ixprime3microx

prime3 middot

part


part


)eiKmiddotx

prime3vminus

= ix23part

partxmicro3

1

x23eiKmiddotx

prime3vminus (C2)


parth


part

partxmicro2

(1

x22h

)minus x23

part

partxmicro3

(1

x23h

)minus x24

part

partxmicro4

(1

x24h

)(C3)

ndash 33 ndash

JHEP03(2016)165


x22x23x

24

(part

partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)[h

x22x23x

24

]= 0 (C4)


x1micropart

partx1νminus x1ν

part

partx1micro

)h= F1(Mmicroνe









x22x23x

24

(x1micro

part

partx1νminus x1ν

part


part

partx2νminus x2ν

part


part

partx3νminus x3ν

part

partx3micro

+x4micropart

partx4νminus x4ν

part

partx4micro

)[h

x22x23x

24

]= 0 (C6)


consider(x1 middot

part

partx1+ 3

)h = F1(e






x22x23x

24

(x1 middot

part

partx1+ x2 middot

part

partx2+ x3 middot

part

partx3+ x4 middot

part

partx4+ 12

)[h

x22x23x

24

]= 0 (C8)


h

x22x23x

24

=A




we have




ndash 34 ndash

JHEP03(2016)165





infinsumnm=0

im(minus1)n




where







minus (C14)

we find


and

infinsumn=0

(minus1)n




1


(C16)



=infinsum

n1n3n4=0

1



slot we get

α(P middot x2) =1


ndash 35 ndash

JHEP03(2016)165




=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)


=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)

(n1 + n3 + n4)



=infinsumn1

(minus1)n1




timesinfinsumn3

(minus1)n3




timesinfinsumn4

(minus1)n4





=1


1


1


=x22x

33x

24






References














ndash 36 ndash

JHEP03(2016)165










(2003) 013 [hep-th0212208] [INSPIRE]




(2014) 302 [arXiv14036646] [INSPIRE]






















(2015) 006 [arXiv150208005] [INSPIRE]






ndash 37 ndash

JHEP03(2016)165



[INSPIRE]

















(2015) 29FT01 [arXiv150500963] [INSPIRE]








[INSPIRE]














ndash 38 ndash

JHEP03(2016)165















ndash 39 ndash

Introduction























JHEP03(2016)165






F1 =1

λlog

((1 + ρs)


)(419)



sk

u2k+1[Qk(u) log(1 + u)]+ (420)



Qk(u) =

ksuml=0

bl(k)ul (421)

where

bl(k) =

(k

k minus l

)(2k + l

k minus l

)(422)








equivariant maps


I = I1 + I2 + I3 + I4 + I5



I(1) = (IS11 + IS21)

I(2) = (IS22 + IS32 + IL32)

I(3) = (IL33 + IS33 + IS43)

I(4) = (IS45 + IS55) (52)

ndash 17 ndash

JHEP03(2016)165




I = I(1) + I(2) + I(3) + I(4) (53)



jI(i) = minus2δij

prodk 6=j

1

(xk minus xi)2(54)









I(1) =sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

r22

Y l3m3

(xprime3)

r23

Y l4m4

(xprime4)



(l1 + l2 + l3 + l4 + 4)(minusl1 + l2 + l3 + l4 + 2)(55)




part2


(r21)A = 4A(A+ 1)(r21)Aminus1

part


part

partxmicro1Y l1m1

(x1)part


21)Aminus1Y l1

m1(x1) (56)


part2


I(1) =sumlimi


r22r23r

24

Y l1m1

(x1)Yl2m2

(xprime2)Yl3m3

(xprime3)Yl4m4


(57)


1


1

r22r23r

24

sumlimi

Y l2m2

(x1)Yl3m3

(x1)Yl4m4

(x1)Yl2m2

(xprime2)Yl3m3

(xprime3)Yl4m4

(xprime4)

=(l1 + 1)rl2+l3+l41

r22r23r

24

sumlimi

Cm1m2m3m4l1l2l3l4

Y l1m1

(x)Y l2m2

(xprime2)Yl3m3

(xprime3)Yl4m4

(xprime4) (58)

ndash 18 ndash

JHEP03(2016)165














I(2) =sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


1

l2+l3+l4minusl1+2

δ(l1 + l2 l3 + l4)

r23r24

+sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


1

r23r24

times



]


part

partxmicro2Y l2m2

(x2)part


22)Aminus1Y l2

m2(x2)

part

partxmicro2logr2 =

xmicro2r22

part

partxmicro2

part

partxmicro2logr2 =

2

r22(59)

We find

part2




r23r24

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


(510)


minus 2







ndash 19 ndash

JHEP03(2016)165


equivariant map


such that



x22x23x

24








23x

24


prime2)f(x1 x

prime3)f(x1 x

prime4)

(515)



1


(516)







part2



prime2)f(x1 x

prime3)f(x1 x

prime4) (518)


24I

(1)(x1 x2 x3 x4)


x22x23x

24 I(1)(x1 x2 x3 x4)rarr F1(e


While x23x24I








ndash 20 ndash

JHEP03(2016)165

implies that (part

partxmicro1+ x22

part

partxmicro2

1

x22+ x23

part

partxmicro3

1

x23+ x24

part

partxmicro4

1

x24

)F1 = 0 (521)


24I



24(I

(1)+I(2)+I(3)+I(4)) = x22x23x



partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)I = 0 (522)





23x

24I(x1 x2 x3 x4) =

x22x23x

24

x213x224

Φ(s t)

(523)



with


minus)=xprime21 x

prime22 xprime23


The x21x22x

23I













〈φ(x1)φ3(x2)φ(x3)φ(x4)〉 (528)




ndash 21 ndash

JHEP03(2016)165



twisted co-product










23x

24



















prime22 xprime23






ndash 22 ndash

JHEP03(2016)165





I2 =

intd4x5d

4x6xminus256

x215x225x

245x

226x

246x

236

(61)






















ndash 23 ndash

JHEP03(2016)165
































ndash 24 ndash

JHEP03(2016)165































Acknowledgments








this work was done

ndash 25 ndash

JHEP03(2016)165








1

2π2

intdS3Y l1

m1(x)Y l2

m2(x) = gm1m2

δl1l2

(l1 + 1)(A1)




gm1m2 = δm1m2


m2(x) = Y l

m1(x)

Y ml (x) = Y l

m(x) (A2)






(mLmR)

gm1m2 = δm1minusm2

Y ml (x) = Y l

minusm(x) (A3)



ables are

z = x1 + ix2 harr(

1

2

1

2


(minus1

2minus1

2

)w = x3 + ix4 harr

(1

2minus1

2


(minus1

21

2

)(A4)

We find that

Y l(l2l2) = (x1 + ix2)

l = zl (A5)

has normalization

1

2π2

intdS3(Y l

( l2 l2))lowast Y l

( l2 l2)

=1

(l + 1)(A6)


ndash 26 ndash

JHEP03(2016)165




(minusi)l

2llY lm(x)Y m

l (P )v+ (A7)


η(Y l1m1

(P )v+ Y m2l2

(K)vminus)

= δl1l2δm1m222l1(l1)2 (A8)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)= δl1l2δm1m2 (A9)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)=

1

2π2

intS3

Y m2l2

(xprime)


m1(x)) (A10)





Y lm(xprime)

xprime2


m(x) (A12)


η = f η


1


summ

1

y2Y lm(x)Y m


=infinsuml=0

summ

1

x2Y lm(xprime)Y m

l (y) if |x| gt |y| (A13)






ndash 27 ndash

JHEP03(2016)165


C2 = minus1

2LMNL


2MpqMqp (A14)






(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)eminusiP middotxv+

=

(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)ipart





Mpq = xppart

partxqminus xq

part

partxp(A16)


1

2MpqMqp(r

AY lm(x)) =

rA

2MpqMqpY


m(x) (A17)





m(x) (A19)

Thus we have

C2 rAY l




m(x) (A21)





F1 =1

λlog

((1 + ρs)


)(B1)

ndash 28 ndash

JHEP03(2016)165


of s to obtain

F1 =log(1+u)

u+

s


s2

u5(3u(2+u)minus 6 + 6u+ u2) log(1+u)

)+s3


)(B2)


F1 =

(minus1 +

u

2minus u2

3+u3


)+ s

(minus1

6+u

6minus 3u2


)+ s2

(minus1

30+

u

20minus 2u2


)+s3

(minus1

140+

u

70minus 5u2



The term at sk is

sk









in Pk(u) is

(2k)

(k)2(B6)


Qk(u) =ksuml=0

blul (B7)


ksuml=0

bl(minus1)l

(iminus l)= 0 (B8)




b1 =5b02 b2 =

20b09 b3 =

5b06 b4 =

5b042 b5 =

b0252

(B9)

ndash 29 ndash

JHEP03(2016)165



ksuml=0

MIlbl = 0 (B10)

where

MIl =(minus1)l




MIlbl (B12)



bl =sumI




bIb0 = 1 16bIb0 = 32 35 120bIb0 = 2 97 27 170

bIb0 =

5

2

20

95

6

5

42

1

252

bIb0 =

3

75

22

20

11

5

11

1

22

1

924

bIb0 =

7

2

63

13175

52

175

143

63

286

7

429

1

3432

(B14)



b1(k)b0 = k2

bk(k)b0 =k2

(2k)

bkminus1(k)b0 =k2

(2k)k(k + 1)

bkminus2(k)b0 =k2

(2k)

(k minus 1)k(k + 1)(k + 2)

4

bkminus3(k)b0 =k2

(2k)

(k

3

)(k + 3

3

)bkminusm(k)b0 =

k2

(2k)

(k

m

)(k +m

m

)(B15)

ndash 30 ndash

JHEP03(2016)165


so that

we have

b1(k) = k2(2k)

k2

bk(k) = 1



4

bkminus3(k) =

(k

3

)(k + 3

3

)bkminusm(k) =

(k

m

)(k +m

m

)(B16)


Pk(u) =

ksumi=1

aiuiminus1 (B17)



bl(minus1)iminusl

(iminus l)= 0 (B18)




k = 2 ~a = a1 ak = 6 3

k = 3 ~a =1

3times 60 60 11

k = 4 ~a =5

6times 84 126 52 5

k = 5 ~a =1

30times 7560 15120 9870 2310 137

k = 6 ~a =7

10times 13720 3300 2960 1140 174 7 (B19)

which leads to

a1(k) =(2k)

(k)2

ak(k) = 2h(k) = 2

ksuml=1

1

l(B20)




ndash 31 ndash

JHEP03(2016)165


grals






infinsumli=0

summi

infinsumt=0

sump


(x1)Yl2minusm2

(x2)Yl3m3

(xprime3)Yl4m4



t+ 1

=x23x

24

x213x224

1

λlog

((1 + ρs)


)(B21)



C l1 l2 lm1m2 m1+m2

=

langl12

l22

mL12

mL22

∣∣∣∣∣ l2

mL1 +mL2

2

ranglangl12

l22

mR12

mR22

∣∣∣∣∣ l2

mR1 +mR22






Yn12

n12n12

(x1)Yn22

n22n22



(x4) = zn11 zn2

2 zn33 zn4

4 (B22)


zn11 zn2

2 zn33 zn4


note that





that zn11 zn2

2 zn33 zn4


Yn12

n12n1minus2

2

(x1)Yn22

n2minus22

n22

(x2)Yn32


2

(x3)Yn42

minusn4minus22

minusn42

(x4)

=radicn1n2n3n4z

n1minus11 w1z

n2minus12 w2z

n3minus13 w3z

n4minus14 w4 (B24)


coefficients

Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B25)

Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B26)

ndash 32 ndash

JHEP03(2016)165


Cn12n22n1+n2

2n12n2minus2

2n1+n2minus2

2

=

radicn2

n1 + n2(B27)

Cn12n22n1+n2minus2

2n12n2minus2

2n1+n2minus2

2

= minusradic

n1n1 + n2

(B28)



n2minus12 w2z

n3minus13 w3z

n4minus14 w4 is

radicn1n2n3n4

[Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2 + 1

+Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2

]=


(B29)




section 5






parth








prime4vminus) (C1)



(2ixprime3microx

prime3 middot

part


part


)eiKmiddotx

prime3vminus

= ix23part

partxmicro3

1

x23eiKmiddotx

prime3vminus (C2)


parth


part

partxmicro2

(1

x22h

)minus x23

part

partxmicro3

(1

x23h

)minus x24

part

partxmicro4

(1

x24h

)(C3)

ndash 33 ndash

JHEP03(2016)165


x22x23x

24

(part

partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)[h

x22x23x

24

]= 0 (C4)


x1micropart

partx1νminus x1ν

part

partx1micro

)h= F1(Mmicroνe









x22x23x

24

(x1micro

part

partx1νminus x1ν

part


part

partx2νminus x2ν

part


part

partx3νminus x3ν

part

partx3micro

+x4micropart

partx4νminus x4ν

part

partx4micro

)[h

x22x23x

24

]= 0 (C6)


consider(x1 middot

part

partx1+ 3

)h = F1(e






x22x23x

24

(x1 middot

part

partx1+ x2 middot

part

partx2+ x3 middot

part

partx3+ x4 middot

part

partx4+ 12

)[h

x22x23x

24

]= 0 (C8)


h

x22x23x

24

=A




we have




ndash 34 ndash

JHEP03(2016)165





infinsumnm=0

im(minus1)n




where







minus (C14)

we find


and

infinsumn=0

(minus1)n




1


(C16)



=infinsum

n1n3n4=0

1



slot we get

α(P middot x2) =1


ndash 35 ndash

JHEP03(2016)165




=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)


=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)

(n1 + n3 + n4)



=infinsumn1

(minus1)n1




timesinfinsumn3

(minus1)n3




timesinfinsumn4

(minus1)n4





=1


1


1


=x22x

33x

24






References














ndash 36 ndash

JHEP03(2016)165










(2003) 013 [hep-th0212208] [INSPIRE]




(2014) 302 [arXiv14036646] [INSPIRE]






















(2015) 006 [arXiv150208005] [INSPIRE]






ndash 37 ndash

JHEP03(2016)165



[INSPIRE]

















(2015) 29FT01 [arXiv150500963] [INSPIRE]








[INSPIRE]














ndash 38 ndash

JHEP03(2016)165















ndash 39 ndash

Introduction























JHEP03(2016)165




I = I(1) + I(2) + I(3) + I(4) (53)



jI(i) = minus2δij

prodk 6=j

1

(xk minus xi)2(54)









I(1) =sumlimi

Y l1m1

(x1)Y l2m2

(xprime2)

r22

Y l3m3

(xprime3)

r23

Y l4m4

(xprime4)



(l1 + l2 + l3 + l4 + 4)(minusl1 + l2 + l3 + l4 + 2)(55)




part2


(r21)A = 4A(A+ 1)(r21)Aminus1

part


part

partxmicro1Y l1m1

(x1)part


21)Aminus1Y l1

m1(x1) (56)


part2


I(1) =sumlimi


r22r23r

24

Y l1m1

(x1)Yl2m2

(xprime2)Yl3m3

(xprime3)Yl4m4


(57)


1


1

r22r23r

24

sumlimi

Y l2m2

(x1)Yl3m3

(x1)Yl4m4

(x1)Yl2m2

(xprime2)Yl3m3

(xprime3)Yl4m4

(xprime4)

=(l1 + 1)rl2+l3+l41

r22r23r

24

sumlimi

Cm1m2m3m4l1l2l3l4

Y l1m1

(x)Y l2m2

(xprime2)Yl3m3

(xprime3)Yl4m4

(xprime4) (58)

ndash 18 ndash

JHEP03(2016)165














I(2) =sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


1

l2+l3+l4minusl1+2

δ(l1 + l2 l3 + l4)

r23r24

+sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


1

r23r24

times



]


part

partxmicro2Y l2m2

(x2)part


22)Aminus1Y l2

m2(x2)

part

partxmicro2logr2 =

xmicro2r22

part

partxmicro2

part

partxmicro2logr2 =

2

r22(59)

We find

part2




r23r24

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


(510)


minus 2







ndash 19 ndash

JHEP03(2016)165


equivariant map


such that



x22x23x

24








23x

24


prime2)f(x1 x

prime3)f(x1 x

prime4)

(515)



1


(516)







part2



prime2)f(x1 x

prime3)f(x1 x

prime4) (518)


24I

(1)(x1 x2 x3 x4)


x22x23x

24 I(1)(x1 x2 x3 x4)rarr F1(e


While x23x24I








ndash 20 ndash

JHEP03(2016)165

implies that (part

partxmicro1+ x22

part

partxmicro2

1

x22+ x23

part

partxmicro3

1

x23+ x24

part

partxmicro4

1

x24

)F1 = 0 (521)


24I



24(I

(1)+I(2)+I(3)+I(4)) = x22x23x



partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)I = 0 (522)





23x

24I(x1 x2 x3 x4) =

x22x23x

24

x213x224

Φ(s t)

(523)



with


minus)=xprime21 x

prime22 xprime23


The x21x22x

23I













〈φ(x1)φ3(x2)φ(x3)φ(x4)〉 (528)




ndash 21 ndash

JHEP03(2016)165



twisted co-product










23x

24



















prime22 xprime23






ndash 22 ndash

JHEP03(2016)165





I2 =

intd4x5d

4x6xminus256

x215x225x

245x

226x

246x

236

(61)






















ndash 23 ndash

JHEP03(2016)165
































ndash 24 ndash

JHEP03(2016)165































Acknowledgments








this work was done

ndash 25 ndash

JHEP03(2016)165








1

2π2

intdS3Y l1

m1(x)Y l2

m2(x) = gm1m2

δl1l2

(l1 + 1)(A1)




gm1m2 = δm1m2


m2(x) = Y l

m1(x)

Y ml (x) = Y l

m(x) (A2)






(mLmR)

gm1m2 = δm1minusm2

Y ml (x) = Y l

minusm(x) (A3)



ables are

z = x1 + ix2 harr(

1

2

1

2


(minus1

2minus1

2

)w = x3 + ix4 harr

(1

2minus1

2


(minus1

21

2

)(A4)

We find that

Y l(l2l2) = (x1 + ix2)

l = zl (A5)

has normalization

1

2π2

intdS3(Y l

( l2 l2))lowast Y l

( l2 l2)

=1

(l + 1)(A6)


ndash 26 ndash

JHEP03(2016)165




(minusi)l

2llY lm(x)Y m

l (P )v+ (A7)


η(Y l1m1

(P )v+ Y m2l2

(K)vminus)

= δl1l2δm1m222l1(l1)2 (A8)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)= δl1l2δm1m2 (A9)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)=

1

2π2

intS3

Y m2l2

(xprime)


m1(x)) (A10)





Y lm(xprime)

xprime2


m(x) (A12)


η = f η


1


summ

1

y2Y lm(x)Y m


=infinsuml=0

summ

1

x2Y lm(xprime)Y m

l (y) if |x| gt |y| (A13)






ndash 27 ndash

JHEP03(2016)165


C2 = minus1

2LMNL


2MpqMqp (A14)






(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)eminusiP middotxv+

=

(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)ipart





Mpq = xppart

partxqminus xq

part

partxp(A16)


1

2MpqMqp(r

AY lm(x)) =

rA

2MpqMqpY


m(x) (A17)





m(x) (A19)

Thus we have

C2 rAY l




m(x) (A21)





F1 =1

λlog

((1 + ρs)


)(B1)

ndash 28 ndash

JHEP03(2016)165


of s to obtain

F1 =log(1+u)

u+

s


s2

u5(3u(2+u)minus 6 + 6u+ u2) log(1+u)

)+s3


)(B2)


F1 =

(minus1 +

u

2minus u2

3+u3


)+ s

(minus1

6+u

6minus 3u2


)+ s2

(minus1

30+

u

20minus 2u2


)+s3

(minus1

140+

u

70minus 5u2



The term at sk is

sk









in Pk(u) is

(2k)

(k)2(B6)


Qk(u) =ksuml=0

blul (B7)


ksuml=0

bl(minus1)l

(iminus l)= 0 (B8)




b1 =5b02 b2 =

20b09 b3 =

5b06 b4 =

5b042 b5 =

b0252

(B9)

ndash 29 ndash

JHEP03(2016)165



ksuml=0

MIlbl = 0 (B10)

where

MIl =(minus1)l




MIlbl (B12)



bl =sumI




bIb0 = 1 16bIb0 = 32 35 120bIb0 = 2 97 27 170

bIb0 =

5

2

20

95

6

5

42

1

252

bIb0 =

3

75

22

20

11

5

11

1

22

1

924

bIb0 =

7

2

63

13175

52

175

143

63

286

7

429

1

3432

(B14)



b1(k)b0 = k2

bk(k)b0 =k2

(2k)

bkminus1(k)b0 =k2

(2k)k(k + 1)

bkminus2(k)b0 =k2

(2k)

(k minus 1)k(k + 1)(k + 2)

4

bkminus3(k)b0 =k2

(2k)

(k

3

)(k + 3

3

)bkminusm(k)b0 =

k2

(2k)

(k

m

)(k +m

m

)(B15)

ndash 30 ndash

JHEP03(2016)165


so that

we have

b1(k) = k2(2k)

k2

bk(k) = 1



4

bkminus3(k) =

(k

3

)(k + 3

3

)bkminusm(k) =

(k

m

)(k +m

m

)(B16)


Pk(u) =

ksumi=1

aiuiminus1 (B17)



bl(minus1)iminusl

(iminus l)= 0 (B18)




k = 2 ~a = a1 ak = 6 3

k = 3 ~a =1

3times 60 60 11

k = 4 ~a =5

6times 84 126 52 5

k = 5 ~a =1

30times 7560 15120 9870 2310 137

k = 6 ~a =7

10times 13720 3300 2960 1140 174 7 (B19)

which leads to

a1(k) =(2k)

(k)2

ak(k) = 2h(k) = 2

ksuml=1

1

l(B20)




ndash 31 ndash

JHEP03(2016)165


grals






infinsumli=0

summi

infinsumt=0

sump


(x1)Yl2minusm2

(x2)Yl3m3

(xprime3)Yl4m4



t+ 1

=x23x

24

x213x224

1

λlog

((1 + ρs)


)(B21)



C l1 l2 lm1m2 m1+m2

=

langl12

l22

mL12

mL22

∣∣∣∣∣ l2

mL1 +mL2

2

ranglangl12

l22

mR12

mR22

∣∣∣∣∣ l2

mR1 +mR22






Yn12

n12n12

(x1)Yn22

n22n22



(x4) = zn11 zn2

2 zn33 zn4

4 (B22)


zn11 zn2

2 zn33 zn4


note that





that zn11 zn2

2 zn33 zn4


Yn12

n12n1minus2

2

(x1)Yn22

n2minus22

n22

(x2)Yn32


2

(x3)Yn42

minusn4minus22

minusn42

(x4)

=radicn1n2n3n4z

n1minus11 w1z

n2minus12 w2z

n3minus13 w3z

n4minus14 w4 (B24)


coefficients

Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B25)

Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B26)

ndash 32 ndash

JHEP03(2016)165


Cn12n22n1+n2

2n12n2minus2

2n1+n2minus2

2

=

radicn2

n1 + n2(B27)

Cn12n22n1+n2minus2

2n12n2minus2

2n1+n2minus2

2

= minusradic

n1n1 + n2

(B28)



n2minus12 w2z

n3minus13 w3z

n4minus14 w4 is

radicn1n2n3n4

[Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2 + 1

+Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2

]=


(B29)




section 5






parth








prime4vminus) (C1)



(2ixprime3microx

prime3 middot

part


part


)eiKmiddotx

prime3vminus

= ix23part

partxmicro3

1

x23eiKmiddotx

prime3vminus (C2)


parth


part

partxmicro2

(1

x22h

)minus x23

part

partxmicro3

(1

x23h

)minus x24

part

partxmicro4

(1

x24h

)(C3)

ndash 33 ndash

JHEP03(2016)165


x22x23x

24

(part

partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)[h

x22x23x

24

]= 0 (C4)


x1micropart

partx1νminus x1ν

part

partx1micro

)h= F1(Mmicroνe









x22x23x

24

(x1micro

part

partx1νminus x1ν

part


part

partx2νminus x2ν

part


part

partx3νminus x3ν

part

partx3micro

+x4micropart

partx4νminus x4ν

part

partx4micro

)[h

x22x23x

24

]= 0 (C6)


consider(x1 middot

part

partx1+ 3

)h = F1(e






x22x23x

24

(x1 middot

part

partx1+ x2 middot

part

partx2+ x3 middot

part

partx3+ x4 middot

part

partx4+ 12

)[h

x22x23x

24

]= 0 (C8)


h

x22x23x

24

=A




we have




ndash 34 ndash

JHEP03(2016)165





infinsumnm=0

im(minus1)n




where







minus (C14)

we find


and

infinsumn=0

(minus1)n




1


(C16)



=infinsum

n1n3n4=0

1



slot we get

α(P middot x2) =1


ndash 35 ndash

JHEP03(2016)165




=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)


=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)

(n1 + n3 + n4)



=infinsumn1

(minus1)n1




timesinfinsumn3

(minus1)n3




timesinfinsumn4

(minus1)n4





=1


1


1


=x22x

33x

24






References














ndash 36 ndash

JHEP03(2016)165










(2003) 013 [hep-th0212208] [INSPIRE]




(2014) 302 [arXiv14036646] [INSPIRE]






















(2015) 006 [arXiv150208005] [INSPIRE]






ndash 37 ndash

JHEP03(2016)165



[INSPIRE]

















(2015) 29FT01 [arXiv150500963] [INSPIRE]








[INSPIRE]














ndash 38 ndash

JHEP03(2016)165















ndash 39 ndash

Introduction























JHEP03(2016)165














I(2) =sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


1

l2+l3+l4minusl1+2

δ(l1 + l2 l3 + l4)

r23r24

+sumlimi

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


1

r23r24

times



]


part

partxmicro2Y l2m2

(x2)part


22)Aminus1Y l2

m2(x2)

part

partxmicro2logr2 =

xmicro2r22

part

partxmicro2

part

partxmicro2logr2 =

2

r22(59)

We find

part2




r23r24

Y l1m1

(x1)Yl2m2

(x2)Yl3m3

(xprime3)Yl4m4


(510)


minus 2







ndash 19 ndash

JHEP03(2016)165


equivariant map


such that



x22x23x

24








23x

24


prime2)f(x1 x

prime3)f(x1 x

prime4)

(515)



1


(516)







part2



prime2)f(x1 x

prime3)f(x1 x

prime4) (518)


24I

(1)(x1 x2 x3 x4)


x22x23x

24 I(1)(x1 x2 x3 x4)rarr F1(e


While x23x24I








ndash 20 ndash

JHEP03(2016)165

implies that (part

partxmicro1+ x22

part

partxmicro2

1

x22+ x23

part

partxmicro3

1

x23+ x24

part

partxmicro4

1

x24

)F1 = 0 (521)


24I



24(I

(1)+I(2)+I(3)+I(4)) = x22x23x



partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)I = 0 (522)





23x

24I(x1 x2 x3 x4) =

x22x23x

24

x213x224

Φ(s t)

(523)



with


minus)=xprime21 x

prime22 xprime23


The x21x22x

23I













〈φ(x1)φ3(x2)φ(x3)φ(x4)〉 (528)




ndash 21 ndash

JHEP03(2016)165



twisted co-product










23x

24



















prime22 xprime23






ndash 22 ndash

JHEP03(2016)165





I2 =

intd4x5d

4x6xminus256

x215x225x

245x

226x

246x

236

(61)






















ndash 23 ndash

JHEP03(2016)165
































ndash 24 ndash

JHEP03(2016)165































Acknowledgments








this work was done

ndash 25 ndash

JHEP03(2016)165








1

2π2

intdS3Y l1

m1(x)Y l2

m2(x) = gm1m2

δl1l2

(l1 + 1)(A1)




gm1m2 = δm1m2


m2(x) = Y l

m1(x)

Y ml (x) = Y l

m(x) (A2)






(mLmR)

gm1m2 = δm1minusm2

Y ml (x) = Y l

minusm(x) (A3)



ables are

z = x1 + ix2 harr(

1

2

1

2


(minus1

2minus1

2

)w = x3 + ix4 harr

(1

2minus1

2


(minus1

21

2

)(A4)

We find that

Y l(l2l2) = (x1 + ix2)

l = zl (A5)

has normalization

1

2π2

intdS3(Y l

( l2 l2))lowast Y l

( l2 l2)

=1

(l + 1)(A6)


ndash 26 ndash

JHEP03(2016)165




(minusi)l

2llY lm(x)Y m

l (P )v+ (A7)


η(Y l1m1

(P )v+ Y m2l2

(K)vminus)

= δl1l2δm1m222l1(l1)2 (A8)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)= δl1l2δm1m2 (A9)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)=

1

2π2

intS3

Y m2l2

(xprime)


m1(x)) (A10)





Y lm(xprime)

xprime2


m(x) (A12)


η = f η


1


summ

1

y2Y lm(x)Y m


=infinsuml=0

summ

1

x2Y lm(xprime)Y m

l (y) if |x| gt |y| (A13)






ndash 27 ndash

JHEP03(2016)165


C2 = minus1

2LMNL


2MpqMqp (A14)






(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)eminusiP middotxv+

=

(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)ipart





Mpq = xppart

partxqminus xq

part

partxp(A16)


1

2MpqMqp(r

AY lm(x)) =

rA

2MpqMqpY


m(x) (A17)





m(x) (A19)

Thus we have

C2 rAY l




m(x) (A21)





F1 =1

λlog

((1 + ρs)


)(B1)

ndash 28 ndash

JHEP03(2016)165


of s to obtain

F1 =log(1+u)

u+

s


s2

u5(3u(2+u)minus 6 + 6u+ u2) log(1+u)

)+s3


)(B2)


F1 =

(minus1 +

u

2minus u2

3+u3


)+ s

(minus1

6+u

6minus 3u2


)+ s2

(minus1

30+

u

20minus 2u2


)+s3

(minus1

140+

u

70minus 5u2



The term at sk is

sk









in Pk(u) is

(2k)

(k)2(B6)


Qk(u) =ksuml=0

blul (B7)


ksuml=0

bl(minus1)l

(iminus l)= 0 (B8)




b1 =5b02 b2 =

20b09 b3 =

5b06 b4 =

5b042 b5 =

b0252

(B9)

ndash 29 ndash

JHEP03(2016)165



ksuml=0

MIlbl = 0 (B10)

where

MIl =(minus1)l




MIlbl (B12)



bl =sumI




bIb0 = 1 16bIb0 = 32 35 120bIb0 = 2 97 27 170

bIb0 =

5

2

20

95

6

5

42

1

252

bIb0 =

3

75

22

20

11

5

11

1

22

1

924

bIb0 =

7

2

63

13175

52

175

143

63

286

7

429

1

3432

(B14)



b1(k)b0 = k2

bk(k)b0 =k2

(2k)

bkminus1(k)b0 =k2

(2k)k(k + 1)

bkminus2(k)b0 =k2

(2k)

(k minus 1)k(k + 1)(k + 2)

4

bkminus3(k)b0 =k2

(2k)

(k

3

)(k + 3

3

)bkminusm(k)b0 =

k2

(2k)

(k

m

)(k +m

m

)(B15)

ndash 30 ndash

JHEP03(2016)165


so that

we have

b1(k) = k2(2k)

k2

bk(k) = 1



4

bkminus3(k) =

(k

3

)(k + 3

3

)bkminusm(k) =

(k

m

)(k +m

m

)(B16)


Pk(u) =

ksumi=1

aiuiminus1 (B17)



bl(minus1)iminusl

(iminus l)= 0 (B18)




k = 2 ~a = a1 ak = 6 3

k = 3 ~a =1

3times 60 60 11

k = 4 ~a =5

6times 84 126 52 5

k = 5 ~a =1

30times 7560 15120 9870 2310 137

k = 6 ~a =7

10times 13720 3300 2960 1140 174 7 (B19)

which leads to

a1(k) =(2k)

(k)2

ak(k) = 2h(k) = 2

ksuml=1

1

l(B20)




ndash 31 ndash

JHEP03(2016)165


grals






infinsumli=0

summi

infinsumt=0

sump


(x1)Yl2minusm2

(x2)Yl3m3

(xprime3)Yl4m4



t+ 1

=x23x

24

x213x224

1

λlog

((1 + ρs)


)(B21)



C l1 l2 lm1m2 m1+m2

=

langl12

l22

mL12

mL22

∣∣∣∣∣ l2

mL1 +mL2

2

ranglangl12

l22

mR12

mR22

∣∣∣∣∣ l2

mR1 +mR22






Yn12

n12n12

(x1)Yn22

n22n22



(x4) = zn11 zn2

2 zn33 zn4

4 (B22)


zn11 zn2

2 zn33 zn4


note that





that zn11 zn2

2 zn33 zn4


Yn12

n12n1minus2

2

(x1)Yn22

n2minus22

n22

(x2)Yn32


2

(x3)Yn42

minusn4minus22

minusn42

(x4)

=radicn1n2n3n4z

n1minus11 w1z

n2minus12 w2z

n3minus13 w3z

n4minus14 w4 (B24)


coefficients

Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B25)

Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B26)

ndash 32 ndash

JHEP03(2016)165


Cn12n22n1+n2

2n12n2minus2

2n1+n2minus2

2

=

radicn2

n1 + n2(B27)

Cn12n22n1+n2minus2

2n12n2minus2

2n1+n2minus2

2

= minusradic

n1n1 + n2

(B28)



n2minus12 w2z

n3minus13 w3z

n4minus14 w4 is

radicn1n2n3n4

[Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2 + 1

+Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2

]=


(B29)




section 5






parth








prime4vminus) (C1)



(2ixprime3microx

prime3 middot

part


part


)eiKmiddotx

prime3vminus

= ix23part

partxmicro3

1

x23eiKmiddotx

prime3vminus (C2)


parth


part

partxmicro2

(1

x22h

)minus x23

part

partxmicro3

(1

x23h

)minus x24

part

partxmicro4

(1

x24h

)(C3)

ndash 33 ndash

JHEP03(2016)165


x22x23x

24

(part

partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)[h

x22x23x

24

]= 0 (C4)


x1micropart

partx1νminus x1ν

part

partx1micro

)h= F1(Mmicroνe









x22x23x

24

(x1micro

part

partx1νminus x1ν

part


part

partx2νminus x2ν

part


part

partx3νminus x3ν

part

partx3micro

+x4micropart

partx4νminus x4ν

part

partx4micro

)[h

x22x23x

24

]= 0 (C6)


consider(x1 middot

part

partx1+ 3

)h = F1(e






x22x23x

24

(x1 middot

part

partx1+ x2 middot

part

partx2+ x3 middot

part

partx3+ x4 middot

part

partx4+ 12

)[h

x22x23x

24

]= 0 (C8)


h

x22x23x

24

=A




we have




ndash 34 ndash

JHEP03(2016)165





infinsumnm=0

im(minus1)n




where







minus (C14)

we find


and

infinsumn=0

(minus1)n




1


(C16)



=infinsum

n1n3n4=0

1



slot we get

α(P middot x2) =1


ndash 35 ndash

JHEP03(2016)165




=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)


=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)

(n1 + n3 + n4)



=infinsumn1

(minus1)n1




timesinfinsumn3

(minus1)n3




timesinfinsumn4

(minus1)n4





=1


1


1


=x22x

33x

24






References














ndash 36 ndash

JHEP03(2016)165










(2003) 013 [hep-th0212208] [INSPIRE]




(2014) 302 [arXiv14036646] [INSPIRE]






















(2015) 006 [arXiv150208005] [INSPIRE]






ndash 37 ndash

JHEP03(2016)165



[INSPIRE]

















(2015) 29FT01 [arXiv150500963] [INSPIRE]








[INSPIRE]














ndash 38 ndash

JHEP03(2016)165















ndash 39 ndash

Introduction























JHEP03(2016)165


equivariant map


such that



x22x23x

24








23x

24


prime2)f(x1 x

prime3)f(x1 x

prime4)

(515)



1


(516)







part2



prime2)f(x1 x

prime3)f(x1 x

prime4) (518)


24I

(1)(x1 x2 x3 x4)


x22x23x

24 I(1)(x1 x2 x3 x4)rarr F1(e


While x23x24I








ndash 20 ndash

JHEP03(2016)165

implies that (part

partxmicro1+ x22

part

partxmicro2

1

x22+ x23

part

partxmicro3

1

x23+ x24

part

partxmicro4

1

x24

)F1 = 0 (521)


24I



24(I

(1)+I(2)+I(3)+I(4)) = x22x23x



partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)I = 0 (522)





23x

24I(x1 x2 x3 x4) =

x22x23x

24

x213x224

Φ(s t)

(523)



with


minus)=xprime21 x

prime22 xprime23


The x21x22x

23I













〈φ(x1)φ3(x2)φ(x3)φ(x4)〉 (528)




ndash 21 ndash

JHEP03(2016)165



twisted co-product










23x

24



















prime22 xprime23






ndash 22 ndash

JHEP03(2016)165





I2 =

intd4x5d

4x6xminus256

x215x225x

245x

226x

246x

236

(61)






















ndash 23 ndash

JHEP03(2016)165
































ndash 24 ndash

JHEP03(2016)165































Acknowledgments








this work was done

ndash 25 ndash

JHEP03(2016)165








1

2π2

intdS3Y l1

m1(x)Y l2

m2(x) = gm1m2

δl1l2

(l1 + 1)(A1)




gm1m2 = δm1m2


m2(x) = Y l

m1(x)

Y ml (x) = Y l

m(x) (A2)






(mLmR)

gm1m2 = δm1minusm2

Y ml (x) = Y l

minusm(x) (A3)



ables are

z = x1 + ix2 harr(

1

2

1

2


(minus1

2minus1

2

)w = x3 + ix4 harr

(1

2minus1

2


(minus1

21

2

)(A4)

We find that

Y l(l2l2) = (x1 + ix2)

l = zl (A5)

has normalization

1

2π2

intdS3(Y l

( l2 l2))lowast Y l

( l2 l2)

=1

(l + 1)(A6)


ndash 26 ndash

JHEP03(2016)165




(minusi)l

2llY lm(x)Y m

l (P )v+ (A7)


η(Y l1m1

(P )v+ Y m2l2

(K)vminus)

= δl1l2δm1m222l1(l1)2 (A8)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)= δl1l2δm1m2 (A9)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)=

1

2π2

intS3

Y m2l2

(xprime)


m1(x)) (A10)





Y lm(xprime)

xprime2


m(x) (A12)


η = f η


1


summ

1

y2Y lm(x)Y m


=infinsuml=0

summ

1

x2Y lm(xprime)Y m

l (y) if |x| gt |y| (A13)






ndash 27 ndash

JHEP03(2016)165


C2 = minus1

2LMNL


2MpqMqp (A14)






(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)eminusiP middotxv+

=

(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)ipart





Mpq = xppart

partxqminus xq

part

partxp(A16)


1

2MpqMqp(r

AY lm(x)) =

rA

2MpqMqpY


m(x) (A17)





m(x) (A19)

Thus we have

C2 rAY l




m(x) (A21)





F1 =1

λlog

((1 + ρs)


)(B1)

ndash 28 ndash

JHEP03(2016)165


of s to obtain

F1 =log(1+u)

u+

s


s2

u5(3u(2+u)minus 6 + 6u+ u2) log(1+u)

)+s3


)(B2)


F1 =

(minus1 +

u

2minus u2

3+u3


)+ s

(minus1

6+u

6minus 3u2


)+ s2

(minus1

30+

u

20minus 2u2


)+s3

(minus1

140+

u

70minus 5u2



The term at sk is

sk









in Pk(u) is

(2k)

(k)2(B6)


Qk(u) =ksuml=0

blul (B7)


ksuml=0

bl(minus1)l

(iminus l)= 0 (B8)




b1 =5b02 b2 =

20b09 b3 =

5b06 b4 =

5b042 b5 =

b0252

(B9)

ndash 29 ndash

JHEP03(2016)165



ksuml=0

MIlbl = 0 (B10)

where

MIl =(minus1)l




MIlbl (B12)



bl =sumI




bIb0 = 1 16bIb0 = 32 35 120bIb0 = 2 97 27 170

bIb0 =

5

2

20

95

6

5

42

1

252

bIb0 =

3

75

22

20

11

5

11

1

22

1

924

bIb0 =

7

2

63

13175

52

175

143

63

286

7

429

1

3432

(B14)



b1(k)b0 = k2

bk(k)b0 =k2

(2k)

bkminus1(k)b0 =k2

(2k)k(k + 1)

bkminus2(k)b0 =k2

(2k)

(k minus 1)k(k + 1)(k + 2)

4

bkminus3(k)b0 =k2

(2k)

(k

3

)(k + 3

3

)bkminusm(k)b0 =

k2

(2k)

(k

m

)(k +m

m

)(B15)

ndash 30 ndash

JHEP03(2016)165


so that

we have

b1(k) = k2(2k)

k2

bk(k) = 1



4

bkminus3(k) =

(k

3

)(k + 3

3

)bkminusm(k) =

(k

m

)(k +m

m

)(B16)


Pk(u) =

ksumi=1

aiuiminus1 (B17)



bl(minus1)iminusl

(iminus l)= 0 (B18)




k = 2 ~a = a1 ak = 6 3

k = 3 ~a =1

3times 60 60 11

k = 4 ~a =5

6times 84 126 52 5

k = 5 ~a =1

30times 7560 15120 9870 2310 137

k = 6 ~a =7

10times 13720 3300 2960 1140 174 7 (B19)

which leads to

a1(k) =(2k)

(k)2

ak(k) = 2h(k) = 2

ksuml=1

1

l(B20)




ndash 31 ndash

JHEP03(2016)165


grals






infinsumli=0

summi

infinsumt=0

sump


(x1)Yl2minusm2

(x2)Yl3m3

(xprime3)Yl4m4



t+ 1

=x23x

24

x213x224

1

λlog

((1 + ρs)


)(B21)



C l1 l2 lm1m2 m1+m2

=

langl12

l22

mL12

mL22

∣∣∣∣∣ l2

mL1 +mL2

2

ranglangl12

l22

mR12

mR22

∣∣∣∣∣ l2

mR1 +mR22






Yn12

n12n12

(x1)Yn22

n22n22



(x4) = zn11 zn2

2 zn33 zn4

4 (B22)


zn11 zn2

2 zn33 zn4


note that





that zn11 zn2

2 zn33 zn4


Yn12

n12n1minus2

2

(x1)Yn22

n2minus22

n22

(x2)Yn32


2

(x3)Yn42

minusn4minus22

minusn42

(x4)

=radicn1n2n3n4z

n1minus11 w1z

n2minus12 w2z

n3minus13 w3z

n4minus14 w4 (B24)


coefficients

Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B25)

Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B26)

ndash 32 ndash

JHEP03(2016)165


Cn12n22n1+n2

2n12n2minus2

2n1+n2minus2

2

=

radicn2

n1 + n2(B27)

Cn12n22n1+n2minus2

2n12n2minus2

2n1+n2minus2

2

= minusradic

n1n1 + n2

(B28)



n2minus12 w2z

n3minus13 w3z

n4minus14 w4 is

radicn1n2n3n4

[Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2 + 1

+Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2

]=


(B29)




section 5






parth








prime4vminus) (C1)



(2ixprime3microx

prime3 middot

part


part


)eiKmiddotx

prime3vminus

= ix23part

partxmicro3

1

x23eiKmiddotx

prime3vminus (C2)


parth


part

partxmicro2

(1

x22h

)minus x23

part

partxmicro3

(1

x23h

)minus x24

part

partxmicro4

(1

x24h

)(C3)

ndash 33 ndash

JHEP03(2016)165


x22x23x

24

(part

partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)[h

x22x23x

24

]= 0 (C4)


x1micropart

partx1νminus x1ν

part

partx1micro

)h= F1(Mmicroνe









x22x23x

24

(x1micro

part

partx1νminus x1ν

part


part

partx2νminus x2ν

part


part

partx3νminus x3ν

part

partx3micro

+x4micropart

partx4νminus x4ν

part

partx4micro

)[h

x22x23x

24

]= 0 (C6)


consider(x1 middot

part

partx1+ 3

)h = F1(e






x22x23x

24

(x1 middot

part

partx1+ x2 middot

part

partx2+ x3 middot

part

partx3+ x4 middot

part

partx4+ 12

)[h

x22x23x

24

]= 0 (C8)


h

x22x23x

24

=A




we have




ndash 34 ndash

JHEP03(2016)165





infinsumnm=0

im(minus1)n




where







minus (C14)

we find


and

infinsumn=0

(minus1)n




1


(C16)



=infinsum

n1n3n4=0

1



slot we get

α(P middot x2) =1


ndash 35 ndash

JHEP03(2016)165




=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)


=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)

(n1 + n3 + n4)



=infinsumn1

(minus1)n1




timesinfinsumn3

(minus1)n3




timesinfinsumn4

(minus1)n4





=1


1


1


=x22x

33x

24






References














ndash 36 ndash

JHEP03(2016)165










(2003) 013 [hep-th0212208] [INSPIRE]




(2014) 302 [arXiv14036646] [INSPIRE]






















(2015) 006 [arXiv150208005] [INSPIRE]






ndash 37 ndash

JHEP03(2016)165



[INSPIRE]

















(2015) 29FT01 [arXiv150500963] [INSPIRE]








[INSPIRE]














ndash 38 ndash

JHEP03(2016)165















ndash 39 ndash

Introduction























JHEP03(2016)165

implies that (part

partxmicro1+ x22

part

partxmicro2

1

x22+ x23

part

partxmicro3

1

x23+ x24

part

partxmicro4

1

x24

)F1 = 0 (521)


24I



24(I

(1)+I(2)+I(3)+I(4)) = x22x23x



partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)I = 0 (522)





23x

24I(x1 x2 x3 x4) =

x22x23x

24

x213x224

Φ(s t)

(523)



with


minus)=xprime21 x

prime22 xprime23


The x21x22x

23I













〈φ(x1)φ3(x2)φ(x3)φ(x4)〉 (528)




ndash 21 ndash

JHEP03(2016)165



twisted co-product










23x

24



















prime22 xprime23






ndash 22 ndash

JHEP03(2016)165





I2 =

intd4x5d

4x6xminus256

x215x225x

245x

226x

246x

236

(61)






















ndash 23 ndash

JHEP03(2016)165
































ndash 24 ndash

JHEP03(2016)165































Acknowledgments








this work was done

ndash 25 ndash

JHEP03(2016)165








1

2π2

intdS3Y l1

m1(x)Y l2

m2(x) = gm1m2

δl1l2

(l1 + 1)(A1)




gm1m2 = δm1m2


m2(x) = Y l

m1(x)

Y ml (x) = Y l

m(x) (A2)






(mLmR)

gm1m2 = δm1minusm2

Y ml (x) = Y l

minusm(x) (A3)



ables are

z = x1 + ix2 harr(

1

2

1

2


(minus1

2minus1

2

)w = x3 + ix4 harr

(1

2minus1

2


(minus1

21

2

)(A4)

We find that

Y l(l2l2) = (x1 + ix2)

l = zl (A5)

has normalization

1

2π2

intdS3(Y l

( l2 l2))lowast Y l

( l2 l2)

=1

(l + 1)(A6)


ndash 26 ndash

JHEP03(2016)165




(minusi)l

2llY lm(x)Y m

l (P )v+ (A7)


η(Y l1m1

(P )v+ Y m2l2

(K)vminus)

= δl1l2δm1m222l1(l1)2 (A8)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)= δl1l2δm1m2 (A9)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)=

1

2π2

intS3

Y m2l2

(xprime)


m1(x)) (A10)





Y lm(xprime)

xprime2


m(x) (A12)


η = f η


1


summ

1

y2Y lm(x)Y m


=infinsuml=0

summ

1

x2Y lm(xprime)Y m

l (y) if |x| gt |y| (A13)






ndash 27 ndash

JHEP03(2016)165


C2 = minus1

2LMNL


2MpqMqp (A14)






(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)eminusiP middotxv+

=

(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)ipart





Mpq = xppart

partxqminus xq

part

partxp(A16)


1

2MpqMqp(r

AY lm(x)) =

rA

2MpqMqpY


m(x) (A17)





m(x) (A19)

Thus we have

C2 rAY l




m(x) (A21)





F1 =1

λlog

((1 + ρs)


)(B1)

ndash 28 ndash

JHEP03(2016)165


of s to obtain

F1 =log(1+u)

u+

s


s2

u5(3u(2+u)minus 6 + 6u+ u2) log(1+u)

)+s3


)(B2)


F1 =

(minus1 +

u

2minus u2

3+u3


)+ s

(minus1

6+u

6minus 3u2


)+ s2

(minus1

30+

u

20minus 2u2


)+s3

(minus1

140+

u

70minus 5u2



The term at sk is

sk









in Pk(u) is

(2k)

(k)2(B6)


Qk(u) =ksuml=0

blul (B7)


ksuml=0

bl(minus1)l

(iminus l)= 0 (B8)




b1 =5b02 b2 =

20b09 b3 =

5b06 b4 =

5b042 b5 =

b0252

(B9)

ndash 29 ndash

JHEP03(2016)165



ksuml=0

MIlbl = 0 (B10)

where

MIl =(minus1)l




MIlbl (B12)



bl =sumI




bIb0 = 1 16bIb0 = 32 35 120bIb0 = 2 97 27 170

bIb0 =

5

2

20

95

6

5

42

1

252

bIb0 =

3

75

22

20

11

5

11

1

22

1

924

bIb0 =

7

2

63

13175

52

175

143

63

286

7

429

1

3432

(B14)



b1(k)b0 = k2

bk(k)b0 =k2

(2k)

bkminus1(k)b0 =k2

(2k)k(k + 1)

bkminus2(k)b0 =k2

(2k)

(k minus 1)k(k + 1)(k + 2)

4

bkminus3(k)b0 =k2

(2k)

(k

3

)(k + 3

3

)bkminusm(k)b0 =

k2

(2k)

(k

m

)(k +m

m

)(B15)

ndash 30 ndash

JHEP03(2016)165


so that

we have

b1(k) = k2(2k)

k2

bk(k) = 1



4

bkminus3(k) =

(k

3

)(k + 3

3

)bkminusm(k) =

(k

m

)(k +m

m

)(B16)


Pk(u) =

ksumi=1

aiuiminus1 (B17)



bl(minus1)iminusl

(iminus l)= 0 (B18)




k = 2 ~a = a1 ak = 6 3

k = 3 ~a =1

3times 60 60 11

k = 4 ~a =5

6times 84 126 52 5

k = 5 ~a =1

30times 7560 15120 9870 2310 137

k = 6 ~a =7

10times 13720 3300 2960 1140 174 7 (B19)

which leads to

a1(k) =(2k)

(k)2

ak(k) = 2h(k) = 2

ksuml=1

1

l(B20)




ndash 31 ndash

JHEP03(2016)165


grals






infinsumli=0

summi

infinsumt=0

sump


(x1)Yl2minusm2

(x2)Yl3m3

(xprime3)Yl4m4



t+ 1

=x23x

24

x213x224

1

λlog

((1 + ρs)


)(B21)



C l1 l2 lm1m2 m1+m2

=

langl12

l22

mL12

mL22

∣∣∣∣∣ l2

mL1 +mL2

2

ranglangl12

l22

mR12

mR22

∣∣∣∣∣ l2

mR1 +mR22






Yn12

n12n12

(x1)Yn22

n22n22



(x4) = zn11 zn2

2 zn33 zn4

4 (B22)


zn11 zn2

2 zn33 zn4


note that





that zn11 zn2

2 zn33 zn4


Yn12

n12n1minus2

2

(x1)Yn22

n2minus22

n22

(x2)Yn32


2

(x3)Yn42

minusn4minus22

minusn42

(x4)

=radicn1n2n3n4z

n1minus11 w1z

n2minus12 w2z

n3minus13 w3z

n4minus14 w4 (B24)


coefficients

Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B25)

Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B26)

ndash 32 ndash

JHEP03(2016)165


Cn12n22n1+n2

2n12n2minus2

2n1+n2minus2

2

=

radicn2

n1 + n2(B27)

Cn12n22n1+n2minus2

2n12n2minus2

2n1+n2minus2

2

= minusradic

n1n1 + n2

(B28)



n2minus12 w2z

n3minus13 w3z

n4minus14 w4 is

radicn1n2n3n4

[Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2 + 1

+Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2

]=


(B29)




section 5






parth








prime4vminus) (C1)



(2ixprime3microx

prime3 middot

part


part


)eiKmiddotx

prime3vminus

= ix23part

partxmicro3

1

x23eiKmiddotx

prime3vminus (C2)


parth


part

partxmicro2

(1

x22h

)minus x23

part

partxmicro3

(1

x23h

)minus x24

part

partxmicro4

(1

x24h

)(C3)

ndash 33 ndash

JHEP03(2016)165


x22x23x

24

(part

partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)[h

x22x23x

24

]= 0 (C4)


x1micropart

partx1νminus x1ν

part

partx1micro

)h= F1(Mmicroνe









x22x23x

24

(x1micro

part

partx1νminus x1ν

part


part

partx2νminus x2ν

part


part

partx3νminus x3ν

part

partx3micro

+x4micropart

partx4νminus x4ν

part

partx4micro

)[h

x22x23x

24

]= 0 (C6)


consider(x1 middot

part

partx1+ 3

)h = F1(e






x22x23x

24

(x1 middot

part

partx1+ x2 middot

part

partx2+ x3 middot

part

partx3+ x4 middot

part

partx4+ 12

)[h

x22x23x

24

]= 0 (C8)


h

x22x23x

24

=A




we have




ndash 34 ndash

JHEP03(2016)165





infinsumnm=0

im(minus1)n




where







minus (C14)

we find


and

infinsumn=0

(minus1)n




1


(C16)



=infinsum

n1n3n4=0

1



slot we get

α(P middot x2) =1


ndash 35 ndash

JHEP03(2016)165




=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)


=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)

(n1 + n3 + n4)



=infinsumn1

(minus1)n1




timesinfinsumn3

(minus1)n3




timesinfinsumn4

(minus1)n4





=1


1


1


=x22x

33x

24






References














ndash 36 ndash

JHEP03(2016)165










(2003) 013 [hep-th0212208] [INSPIRE]




(2014) 302 [arXiv14036646] [INSPIRE]






















(2015) 006 [arXiv150208005] [INSPIRE]






ndash 37 ndash

JHEP03(2016)165



[INSPIRE]

















(2015) 29FT01 [arXiv150500963] [INSPIRE]








[INSPIRE]














ndash 38 ndash

JHEP03(2016)165















ndash 39 ndash

Introduction























JHEP03(2016)165



twisted co-product










23x

24



















prime22 xprime23






ndash 22 ndash

JHEP03(2016)165





I2 =

intd4x5d

4x6xminus256

x215x225x

245x

226x

246x

236

(61)






















ndash 23 ndash

JHEP03(2016)165
































ndash 24 ndash

JHEP03(2016)165































Acknowledgments








this work was done

ndash 25 ndash

JHEP03(2016)165








1

2π2

intdS3Y l1

m1(x)Y l2

m2(x) = gm1m2

δl1l2

(l1 + 1)(A1)




gm1m2 = δm1m2


m2(x) = Y l

m1(x)

Y ml (x) = Y l

m(x) (A2)






(mLmR)

gm1m2 = δm1minusm2

Y ml (x) = Y l

minusm(x) (A3)



ables are

z = x1 + ix2 harr(

1

2

1

2


(minus1

2minus1

2

)w = x3 + ix4 harr

(1

2minus1

2


(minus1

21

2

)(A4)

We find that

Y l(l2l2) = (x1 + ix2)

l = zl (A5)

has normalization

1

2π2

intdS3(Y l

( l2 l2))lowast Y l

( l2 l2)

=1

(l + 1)(A6)


ndash 26 ndash

JHEP03(2016)165




(minusi)l

2llY lm(x)Y m

l (P )v+ (A7)


η(Y l1m1

(P )v+ Y m2l2

(K)vminus)

= δl1l2δm1m222l1(l1)2 (A8)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)= δl1l2δm1m2 (A9)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)=

1

2π2

intS3

Y m2l2

(xprime)


m1(x)) (A10)





Y lm(xprime)

xprime2


m(x) (A12)


η = f η


1


summ

1

y2Y lm(x)Y m


=infinsuml=0

summ

1

x2Y lm(xprime)Y m

l (y) if |x| gt |y| (A13)






ndash 27 ndash

JHEP03(2016)165


C2 = minus1

2LMNL


2MpqMqp (A14)






(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)eminusiP middotxv+

=

(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)ipart





Mpq = xppart

partxqminus xq

part

partxp(A16)


1

2MpqMqp(r

AY lm(x)) =

rA

2MpqMqpY


m(x) (A17)





m(x) (A19)

Thus we have

C2 rAY l




m(x) (A21)





F1 =1

λlog

((1 + ρs)


)(B1)

ndash 28 ndash

JHEP03(2016)165


of s to obtain

F1 =log(1+u)

u+

s


s2

u5(3u(2+u)minus 6 + 6u+ u2) log(1+u)

)+s3


)(B2)


F1 =

(minus1 +

u

2minus u2

3+u3


)+ s

(minus1

6+u

6minus 3u2


)+ s2

(minus1

30+

u

20minus 2u2


)+s3

(minus1

140+

u

70minus 5u2



The term at sk is

sk









in Pk(u) is

(2k)

(k)2(B6)


Qk(u) =ksuml=0

blul (B7)


ksuml=0

bl(minus1)l

(iminus l)= 0 (B8)




b1 =5b02 b2 =

20b09 b3 =

5b06 b4 =

5b042 b5 =

b0252

(B9)

ndash 29 ndash

JHEP03(2016)165



ksuml=0

MIlbl = 0 (B10)

where

MIl =(minus1)l




MIlbl (B12)



bl =sumI




bIb0 = 1 16bIb0 = 32 35 120bIb0 = 2 97 27 170

bIb0 =

5

2

20

95

6

5

42

1

252

bIb0 =

3

75

22

20

11

5

11

1

22

1

924

bIb0 =

7

2

63

13175

52

175

143

63

286

7

429

1

3432

(B14)



b1(k)b0 = k2

bk(k)b0 =k2

(2k)

bkminus1(k)b0 =k2

(2k)k(k + 1)

bkminus2(k)b0 =k2

(2k)

(k minus 1)k(k + 1)(k + 2)

4

bkminus3(k)b0 =k2

(2k)

(k

3

)(k + 3

3

)bkminusm(k)b0 =

k2

(2k)

(k

m

)(k +m

m

)(B15)

ndash 30 ndash

JHEP03(2016)165


so that

we have

b1(k) = k2(2k)

k2

bk(k) = 1



4

bkminus3(k) =

(k

3

)(k + 3

3

)bkminusm(k) =

(k

m

)(k +m

m

)(B16)


Pk(u) =

ksumi=1

aiuiminus1 (B17)



bl(minus1)iminusl

(iminus l)= 0 (B18)




k = 2 ~a = a1 ak = 6 3

k = 3 ~a =1

3times 60 60 11

k = 4 ~a =5

6times 84 126 52 5

k = 5 ~a =1

30times 7560 15120 9870 2310 137

k = 6 ~a =7

10times 13720 3300 2960 1140 174 7 (B19)

which leads to

a1(k) =(2k)

(k)2

ak(k) = 2h(k) = 2

ksuml=1

1

l(B20)




ndash 31 ndash

JHEP03(2016)165


grals






infinsumli=0

summi

infinsumt=0

sump


(x1)Yl2minusm2

(x2)Yl3m3

(xprime3)Yl4m4



t+ 1

=x23x

24

x213x224

1

λlog

((1 + ρs)


)(B21)



C l1 l2 lm1m2 m1+m2

=

langl12

l22

mL12

mL22

∣∣∣∣∣ l2

mL1 +mL2

2

ranglangl12

l22

mR12

mR22

∣∣∣∣∣ l2

mR1 +mR22






Yn12

n12n12

(x1)Yn22

n22n22



(x4) = zn11 zn2

2 zn33 zn4

4 (B22)


zn11 zn2

2 zn33 zn4


note that





that zn11 zn2

2 zn33 zn4


Yn12

n12n1minus2

2

(x1)Yn22

n2minus22

n22

(x2)Yn32


2

(x3)Yn42

minusn4minus22

minusn42

(x4)

=radicn1n2n3n4z

n1minus11 w1z

n2minus12 w2z

n3minus13 w3z

n4minus14 w4 (B24)


coefficients

Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B25)

Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B26)

ndash 32 ndash

JHEP03(2016)165


Cn12n22n1+n2

2n12n2minus2

2n1+n2minus2

2

=

radicn2

n1 + n2(B27)

Cn12n22n1+n2minus2

2n12n2minus2

2n1+n2minus2

2

= minusradic

n1n1 + n2

(B28)



n2minus12 w2z

n3minus13 w3z

n4minus14 w4 is

radicn1n2n3n4

[Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2 + 1

+Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2

]=


(B29)




section 5






parth








prime4vminus) (C1)



(2ixprime3microx

prime3 middot

part


part


)eiKmiddotx

prime3vminus

= ix23part

partxmicro3

1

x23eiKmiddotx

prime3vminus (C2)


parth


part

partxmicro2

(1

x22h

)minus x23

part

partxmicro3

(1

x23h

)minus x24

part

partxmicro4

(1

x24h

)(C3)

ndash 33 ndash

JHEP03(2016)165


x22x23x

24

(part

partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)[h

x22x23x

24

]= 0 (C4)


x1micropart

partx1νminus x1ν

part

partx1micro

)h= F1(Mmicroνe









x22x23x

24

(x1micro

part

partx1νminus x1ν

part


part

partx2νminus x2ν

part


part

partx3νminus x3ν

part

partx3micro

+x4micropart

partx4νminus x4ν

part

partx4micro

)[h

x22x23x

24

]= 0 (C6)


consider(x1 middot

part

partx1+ 3

)h = F1(e






x22x23x

24

(x1 middot

part

partx1+ x2 middot

part

partx2+ x3 middot

part

partx3+ x4 middot

part

partx4+ 12

)[h

x22x23x

24

]= 0 (C8)


h

x22x23x

24

=A




we have




ndash 34 ndash

JHEP03(2016)165





infinsumnm=0

im(minus1)n




where







minus (C14)

we find


and

infinsumn=0

(minus1)n




1


(C16)



=infinsum

n1n3n4=0

1



slot we get

α(P middot x2) =1


ndash 35 ndash

JHEP03(2016)165




=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)


=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)

(n1 + n3 + n4)



=infinsumn1

(minus1)n1




timesinfinsumn3

(minus1)n3




timesinfinsumn4

(minus1)n4





=1


1


1


=x22x

33x

24






References














ndash 36 ndash

JHEP03(2016)165










(2003) 013 [hep-th0212208] [INSPIRE]




(2014) 302 [arXiv14036646] [INSPIRE]






















(2015) 006 [arXiv150208005] [INSPIRE]






ndash 37 ndash

JHEP03(2016)165



[INSPIRE]

















(2015) 29FT01 [arXiv150500963] [INSPIRE]








[INSPIRE]














ndash 38 ndash

JHEP03(2016)165















ndash 39 ndash

Introduction























JHEP03(2016)165





I2 =

intd4x5d

4x6xminus256

x215x225x

245x

226x

246x

236

(61)






















ndash 23 ndash

JHEP03(2016)165
































ndash 24 ndash

JHEP03(2016)165































Acknowledgments








this work was done

ndash 25 ndash

JHEP03(2016)165








1

2π2

intdS3Y l1

m1(x)Y l2

m2(x) = gm1m2

δl1l2

(l1 + 1)(A1)




gm1m2 = δm1m2


m2(x) = Y l

m1(x)

Y ml (x) = Y l

m(x) (A2)






(mLmR)

gm1m2 = δm1minusm2

Y ml (x) = Y l

minusm(x) (A3)



ables are

z = x1 + ix2 harr(

1

2

1

2


(minus1

2minus1

2

)w = x3 + ix4 harr

(1

2minus1

2


(minus1

21

2

)(A4)

We find that

Y l(l2l2) = (x1 + ix2)

l = zl (A5)

has normalization

1

2π2

intdS3(Y l

( l2 l2))lowast Y l

( l2 l2)

=1

(l + 1)(A6)


ndash 26 ndash

JHEP03(2016)165




(minusi)l

2llY lm(x)Y m

l (P )v+ (A7)


η(Y l1m1

(P )v+ Y m2l2

(K)vminus)

= δl1l2δm1m222l1(l1)2 (A8)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)= δl1l2δm1m2 (A9)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)=

1

2π2

intS3

Y m2l2

(xprime)


m1(x)) (A10)





Y lm(xprime)

xprime2


m(x) (A12)


η = f η


1


summ

1

y2Y lm(x)Y m


=infinsuml=0

summ

1

x2Y lm(xprime)Y m

l (y) if |x| gt |y| (A13)






ndash 27 ndash

JHEP03(2016)165


C2 = minus1

2LMNL


2MpqMqp (A14)






(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)eminusiP middotxv+

=

(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)ipart





Mpq = xppart

partxqminus xq

part

partxp(A16)


1

2MpqMqp(r

AY lm(x)) =

rA

2MpqMqpY


m(x) (A17)





m(x) (A19)

Thus we have

C2 rAY l




m(x) (A21)





F1 =1

λlog

((1 + ρs)


)(B1)

ndash 28 ndash

JHEP03(2016)165


of s to obtain

F1 =log(1+u)

u+

s


s2

u5(3u(2+u)minus 6 + 6u+ u2) log(1+u)

)+s3


)(B2)


F1 =

(minus1 +

u

2minus u2

3+u3


)+ s

(minus1

6+u

6minus 3u2


)+ s2

(minus1

30+

u

20minus 2u2


)+s3

(minus1

140+

u

70minus 5u2



The term at sk is

sk









in Pk(u) is

(2k)

(k)2(B6)


Qk(u) =ksuml=0

blul (B7)


ksuml=0

bl(minus1)l

(iminus l)= 0 (B8)




b1 =5b02 b2 =

20b09 b3 =

5b06 b4 =

5b042 b5 =

b0252

(B9)

ndash 29 ndash

JHEP03(2016)165



ksuml=0

MIlbl = 0 (B10)

where

MIl =(minus1)l




MIlbl (B12)



bl =sumI




bIb0 = 1 16bIb0 = 32 35 120bIb0 = 2 97 27 170

bIb0 =

5

2

20

95

6

5

42

1

252

bIb0 =

3

75

22

20

11

5

11

1

22

1

924

bIb0 =

7

2

63

13175

52

175

143

63

286

7

429

1

3432

(B14)



b1(k)b0 = k2

bk(k)b0 =k2

(2k)

bkminus1(k)b0 =k2

(2k)k(k + 1)

bkminus2(k)b0 =k2

(2k)

(k minus 1)k(k + 1)(k + 2)

4

bkminus3(k)b0 =k2

(2k)

(k

3

)(k + 3

3

)bkminusm(k)b0 =

k2

(2k)

(k

m

)(k +m

m

)(B15)

ndash 30 ndash

JHEP03(2016)165


so that

we have

b1(k) = k2(2k)

k2

bk(k) = 1



4

bkminus3(k) =

(k

3

)(k + 3

3

)bkminusm(k) =

(k

m

)(k +m

m

)(B16)


Pk(u) =

ksumi=1

aiuiminus1 (B17)



bl(minus1)iminusl

(iminus l)= 0 (B18)




k = 2 ~a = a1 ak = 6 3

k = 3 ~a =1

3times 60 60 11

k = 4 ~a =5

6times 84 126 52 5

k = 5 ~a =1

30times 7560 15120 9870 2310 137

k = 6 ~a =7

10times 13720 3300 2960 1140 174 7 (B19)

which leads to

a1(k) =(2k)

(k)2

ak(k) = 2h(k) = 2

ksuml=1

1

l(B20)




ndash 31 ndash

JHEP03(2016)165


grals






infinsumli=0

summi

infinsumt=0

sump


(x1)Yl2minusm2

(x2)Yl3m3

(xprime3)Yl4m4



t+ 1

=x23x

24

x213x224

1

λlog

((1 + ρs)


)(B21)



C l1 l2 lm1m2 m1+m2

=

langl12

l22

mL12

mL22

∣∣∣∣∣ l2

mL1 +mL2

2

ranglangl12

l22

mR12

mR22

∣∣∣∣∣ l2

mR1 +mR22






Yn12

n12n12

(x1)Yn22

n22n22



(x4) = zn11 zn2

2 zn33 zn4

4 (B22)


zn11 zn2

2 zn33 zn4


note that





that zn11 zn2

2 zn33 zn4


Yn12

n12n1minus2

2

(x1)Yn22

n2minus22

n22

(x2)Yn32


2

(x3)Yn42

minusn4minus22

minusn42

(x4)

=radicn1n2n3n4z

n1minus11 w1z

n2minus12 w2z

n3minus13 w3z

n4minus14 w4 (B24)


coefficients

Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B25)

Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B26)

ndash 32 ndash

JHEP03(2016)165


Cn12n22n1+n2

2n12n2minus2

2n1+n2minus2

2

=

radicn2

n1 + n2(B27)

Cn12n22n1+n2minus2

2n12n2minus2

2n1+n2minus2

2

= minusradic

n1n1 + n2

(B28)



n2minus12 w2z

n3minus13 w3z

n4minus14 w4 is

radicn1n2n3n4

[Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2 + 1

+Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2

]=


(B29)




section 5






parth








prime4vminus) (C1)



(2ixprime3microx

prime3 middot

part


part


)eiKmiddotx

prime3vminus

= ix23part

partxmicro3

1

x23eiKmiddotx

prime3vminus (C2)


parth


part

partxmicro2

(1

x22h

)minus x23

part

partxmicro3

(1

x23h

)minus x24

part

partxmicro4

(1

x24h

)(C3)

ndash 33 ndash

JHEP03(2016)165


x22x23x

24

(part

partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)[h

x22x23x

24

]= 0 (C4)


x1micropart

partx1νminus x1ν

part

partx1micro

)h= F1(Mmicroνe









x22x23x

24

(x1micro

part

partx1νminus x1ν

part


part

partx2νminus x2ν

part


part

partx3νminus x3ν

part

partx3micro

+x4micropart

partx4νminus x4ν

part

partx4micro

)[h

x22x23x

24

]= 0 (C6)


consider(x1 middot

part

partx1+ 3

)h = F1(e






x22x23x

24

(x1 middot

part

partx1+ x2 middot

part

partx2+ x3 middot

part

partx3+ x4 middot

part

partx4+ 12

)[h

x22x23x

24

]= 0 (C8)


h

x22x23x

24

=A




we have




ndash 34 ndash

JHEP03(2016)165





infinsumnm=0

im(minus1)n




where







minus (C14)

we find


and

infinsumn=0

(minus1)n




1


(C16)



=infinsum

n1n3n4=0

1



slot we get

α(P middot x2) =1


ndash 35 ndash

JHEP03(2016)165




=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)


=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)

(n1 + n3 + n4)



=infinsumn1

(minus1)n1




timesinfinsumn3

(minus1)n3




timesinfinsumn4

(minus1)n4





=1


1


1


=x22x

33x

24






References














ndash 36 ndash

JHEP03(2016)165










(2003) 013 [hep-th0212208] [INSPIRE]




(2014) 302 [arXiv14036646] [INSPIRE]






















(2015) 006 [arXiv150208005] [INSPIRE]






ndash 37 ndash

JHEP03(2016)165



[INSPIRE]

















(2015) 29FT01 [arXiv150500963] [INSPIRE]








[INSPIRE]














ndash 38 ndash

JHEP03(2016)165















ndash 39 ndash

Introduction























JHEP03(2016)165
































ndash 24 ndash

JHEP03(2016)165































Acknowledgments








this work was done

ndash 25 ndash

JHEP03(2016)165








1

2π2

intdS3Y l1

m1(x)Y l2

m2(x) = gm1m2

δl1l2

(l1 + 1)(A1)




gm1m2 = δm1m2


m2(x) = Y l

m1(x)

Y ml (x) = Y l

m(x) (A2)






(mLmR)

gm1m2 = δm1minusm2

Y ml (x) = Y l

minusm(x) (A3)



ables are

z = x1 + ix2 harr(

1

2

1

2


(minus1

2minus1

2

)w = x3 + ix4 harr

(1

2minus1

2


(minus1

21

2

)(A4)

We find that

Y l(l2l2) = (x1 + ix2)

l = zl (A5)

has normalization

1

2π2

intdS3(Y l

( l2 l2))lowast Y l

( l2 l2)

=1

(l + 1)(A6)


ndash 26 ndash

JHEP03(2016)165




(minusi)l

2llY lm(x)Y m

l (P )v+ (A7)


η(Y l1m1

(P )v+ Y m2l2

(K)vminus)

= δl1l2δm1m222l1(l1)2 (A8)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)= δl1l2δm1m2 (A9)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)=

1

2π2

intS3

Y m2l2

(xprime)


m1(x)) (A10)





Y lm(xprime)

xprime2


m(x) (A12)


η = f η


1


summ

1

y2Y lm(x)Y m


=infinsuml=0

summ

1

x2Y lm(xprime)Y m

l (y) if |x| gt |y| (A13)






ndash 27 ndash

JHEP03(2016)165


C2 = minus1

2LMNL


2MpqMqp (A14)






(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)eminusiP middotxv+

=

(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)ipart





Mpq = xppart

partxqminus xq

part

partxp(A16)


1

2MpqMqp(r

AY lm(x)) =

rA

2MpqMqpY


m(x) (A17)





m(x) (A19)

Thus we have

C2 rAY l




m(x) (A21)





F1 =1

λlog

((1 + ρs)


)(B1)

ndash 28 ndash

JHEP03(2016)165


of s to obtain

F1 =log(1+u)

u+

s


s2

u5(3u(2+u)minus 6 + 6u+ u2) log(1+u)

)+s3


)(B2)


F1 =

(minus1 +

u

2minus u2

3+u3


)+ s

(minus1

6+u

6minus 3u2


)+ s2

(minus1

30+

u

20minus 2u2


)+s3

(minus1

140+

u

70minus 5u2



The term at sk is

sk









in Pk(u) is

(2k)

(k)2(B6)


Qk(u) =ksuml=0

blul (B7)


ksuml=0

bl(minus1)l

(iminus l)= 0 (B8)




b1 =5b02 b2 =

20b09 b3 =

5b06 b4 =

5b042 b5 =

b0252

(B9)

ndash 29 ndash

JHEP03(2016)165



ksuml=0

MIlbl = 0 (B10)

where

MIl =(minus1)l




MIlbl (B12)



bl =sumI




bIb0 = 1 16bIb0 = 32 35 120bIb0 = 2 97 27 170

bIb0 =

5

2

20

95

6

5

42

1

252

bIb0 =

3

75

22

20

11

5

11

1

22

1

924

bIb0 =

7

2

63

13175

52

175

143

63

286

7

429

1

3432

(B14)



b1(k)b0 = k2

bk(k)b0 =k2

(2k)

bkminus1(k)b0 =k2

(2k)k(k + 1)

bkminus2(k)b0 =k2

(2k)

(k minus 1)k(k + 1)(k + 2)

4

bkminus3(k)b0 =k2

(2k)

(k

3

)(k + 3

3

)bkminusm(k)b0 =

k2

(2k)

(k

m

)(k +m

m

)(B15)

ndash 30 ndash

JHEP03(2016)165


so that

we have

b1(k) = k2(2k)

k2

bk(k) = 1



4

bkminus3(k) =

(k

3

)(k + 3

3

)bkminusm(k) =

(k

m

)(k +m

m

)(B16)


Pk(u) =

ksumi=1

aiuiminus1 (B17)



bl(minus1)iminusl

(iminus l)= 0 (B18)




k = 2 ~a = a1 ak = 6 3

k = 3 ~a =1

3times 60 60 11

k = 4 ~a =5

6times 84 126 52 5

k = 5 ~a =1

30times 7560 15120 9870 2310 137

k = 6 ~a =7

10times 13720 3300 2960 1140 174 7 (B19)

which leads to

a1(k) =(2k)

(k)2

ak(k) = 2h(k) = 2

ksuml=1

1

l(B20)




ndash 31 ndash

JHEP03(2016)165


grals






infinsumli=0

summi

infinsumt=0

sump


(x1)Yl2minusm2

(x2)Yl3m3

(xprime3)Yl4m4



t+ 1

=x23x

24

x213x224

1

λlog

((1 + ρs)


)(B21)



C l1 l2 lm1m2 m1+m2

=

langl12

l22

mL12

mL22

∣∣∣∣∣ l2

mL1 +mL2

2

ranglangl12

l22

mR12

mR22

∣∣∣∣∣ l2

mR1 +mR22






Yn12

n12n12

(x1)Yn22

n22n22



(x4) = zn11 zn2

2 zn33 zn4

4 (B22)


zn11 zn2

2 zn33 zn4


note that





that zn11 zn2

2 zn33 zn4


Yn12

n12n1minus2

2

(x1)Yn22

n2minus22

n22

(x2)Yn32


2

(x3)Yn42

minusn4minus22

minusn42

(x4)

=radicn1n2n3n4z

n1minus11 w1z

n2minus12 w2z

n3minus13 w3z

n4minus14 w4 (B24)


coefficients

Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B25)

Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B26)

ndash 32 ndash

JHEP03(2016)165


Cn12n22n1+n2

2n12n2minus2

2n1+n2minus2

2

=

radicn2

n1 + n2(B27)

Cn12n22n1+n2minus2

2n12n2minus2

2n1+n2minus2

2

= minusradic

n1n1 + n2

(B28)



n2minus12 w2z

n3minus13 w3z

n4minus14 w4 is

radicn1n2n3n4

[Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2 + 1

+Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2

]=


(B29)




section 5






parth








prime4vminus) (C1)



(2ixprime3microx

prime3 middot

part


part


)eiKmiddotx

prime3vminus

= ix23part

partxmicro3

1

x23eiKmiddotx

prime3vminus (C2)


parth


part

partxmicro2

(1

x22h

)minus x23

part

partxmicro3

(1

x23h

)minus x24

part

partxmicro4

(1

x24h

)(C3)

ndash 33 ndash

JHEP03(2016)165


x22x23x

24

(part

partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)[h

x22x23x

24

]= 0 (C4)


x1micropart

partx1νminus x1ν

part

partx1micro

)h= F1(Mmicroνe









x22x23x

24

(x1micro

part

partx1νminus x1ν

part


part

partx2νminus x2ν

part


part

partx3νminus x3ν

part

partx3micro

+x4micropart

partx4νminus x4ν

part

partx4micro

)[h

x22x23x

24

]= 0 (C6)


consider(x1 middot

part

partx1+ 3

)h = F1(e






x22x23x

24

(x1 middot

part

partx1+ x2 middot

part

partx2+ x3 middot

part

partx3+ x4 middot

part

partx4+ 12

)[h

x22x23x

24

]= 0 (C8)


h

x22x23x

24

=A




we have




ndash 34 ndash

JHEP03(2016)165





infinsumnm=0

im(minus1)n




where







minus (C14)

we find


and

infinsumn=0

(minus1)n




1


(C16)



=infinsum

n1n3n4=0

1



slot we get

α(P middot x2) =1


ndash 35 ndash

JHEP03(2016)165




=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)


=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)

(n1 + n3 + n4)



=infinsumn1

(minus1)n1




timesinfinsumn3

(minus1)n3




timesinfinsumn4

(minus1)n4





=1


1


1


=x22x

33x

24






References














ndash 36 ndash

JHEP03(2016)165










(2003) 013 [hep-th0212208] [INSPIRE]




(2014) 302 [arXiv14036646] [INSPIRE]






















(2015) 006 [arXiv150208005] [INSPIRE]






ndash 37 ndash

JHEP03(2016)165



[INSPIRE]

















(2015) 29FT01 [arXiv150500963] [INSPIRE]








[INSPIRE]














ndash 38 ndash

JHEP03(2016)165















ndash 39 ndash

Introduction























JHEP03(2016)165































Acknowledgments








this work was done

ndash 25 ndash

JHEP03(2016)165








1

2π2

intdS3Y l1

m1(x)Y l2

m2(x) = gm1m2

δl1l2

(l1 + 1)(A1)




gm1m2 = δm1m2


m2(x) = Y l

m1(x)

Y ml (x) = Y l

m(x) (A2)






(mLmR)

gm1m2 = δm1minusm2

Y ml (x) = Y l

minusm(x) (A3)



ables are

z = x1 + ix2 harr(

1

2

1

2


(minus1

2minus1

2

)w = x3 + ix4 harr

(1

2minus1

2


(minus1

21

2

)(A4)

We find that

Y l(l2l2) = (x1 + ix2)

l = zl (A5)

has normalization

1

2π2

intdS3(Y l

( l2 l2))lowast Y l

( l2 l2)

=1

(l + 1)(A6)


ndash 26 ndash

JHEP03(2016)165




(minusi)l

2llY lm(x)Y m

l (P )v+ (A7)


η(Y l1m1

(P )v+ Y m2l2

(K)vminus)

= δl1l2δm1m222l1(l1)2 (A8)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)= δl1l2δm1m2 (A9)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)=

1

2π2

intS3

Y m2l2

(xprime)


m1(x)) (A10)





Y lm(xprime)

xprime2


m(x) (A12)


η = f η


1


summ

1

y2Y lm(x)Y m


=infinsuml=0

summ

1

x2Y lm(xprime)Y m

l (y) if |x| gt |y| (A13)






ndash 27 ndash

JHEP03(2016)165


C2 = minus1

2LMNL


2MpqMqp (A14)






(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)eminusiP middotxv+

=

(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)ipart





Mpq = xppart

partxqminus xq

part

partxp(A16)


1

2MpqMqp(r

AY lm(x)) =

rA

2MpqMqpY


m(x) (A17)





m(x) (A19)

Thus we have

C2 rAY l




m(x) (A21)





F1 =1

λlog

((1 + ρs)


)(B1)

ndash 28 ndash

JHEP03(2016)165


of s to obtain

F1 =log(1+u)

u+

s


s2

u5(3u(2+u)minus 6 + 6u+ u2) log(1+u)

)+s3


)(B2)


F1 =

(minus1 +

u

2minus u2

3+u3


)+ s

(minus1

6+u

6minus 3u2


)+ s2

(minus1

30+

u

20minus 2u2


)+s3

(minus1

140+

u

70minus 5u2



The term at sk is

sk









in Pk(u) is

(2k)

(k)2(B6)


Qk(u) =ksuml=0

blul (B7)


ksuml=0

bl(minus1)l

(iminus l)= 0 (B8)




b1 =5b02 b2 =

20b09 b3 =

5b06 b4 =

5b042 b5 =

b0252

(B9)

ndash 29 ndash

JHEP03(2016)165



ksuml=0

MIlbl = 0 (B10)

where

MIl =(minus1)l




MIlbl (B12)



bl =sumI




bIb0 = 1 16bIb0 = 32 35 120bIb0 = 2 97 27 170

bIb0 =

5

2

20

95

6

5

42

1

252

bIb0 =

3

75

22

20

11

5

11

1

22

1

924

bIb0 =

7

2

63

13175

52

175

143

63

286

7

429

1

3432

(B14)



b1(k)b0 = k2

bk(k)b0 =k2

(2k)

bkminus1(k)b0 =k2

(2k)k(k + 1)

bkminus2(k)b0 =k2

(2k)

(k minus 1)k(k + 1)(k + 2)

4

bkminus3(k)b0 =k2

(2k)

(k

3

)(k + 3

3

)bkminusm(k)b0 =

k2

(2k)

(k

m

)(k +m

m

)(B15)

ndash 30 ndash

JHEP03(2016)165


so that

we have

b1(k) = k2(2k)

k2

bk(k) = 1



4

bkminus3(k) =

(k

3

)(k + 3

3

)bkminusm(k) =

(k

m

)(k +m

m

)(B16)


Pk(u) =

ksumi=1

aiuiminus1 (B17)



bl(minus1)iminusl

(iminus l)= 0 (B18)




k = 2 ~a = a1 ak = 6 3

k = 3 ~a =1

3times 60 60 11

k = 4 ~a =5

6times 84 126 52 5

k = 5 ~a =1

30times 7560 15120 9870 2310 137

k = 6 ~a =7

10times 13720 3300 2960 1140 174 7 (B19)

which leads to

a1(k) =(2k)

(k)2

ak(k) = 2h(k) = 2

ksuml=1

1

l(B20)




ndash 31 ndash

JHEP03(2016)165


grals






infinsumli=0

summi

infinsumt=0

sump


(x1)Yl2minusm2

(x2)Yl3m3

(xprime3)Yl4m4



t+ 1

=x23x

24

x213x224

1

λlog

((1 + ρs)


)(B21)



C l1 l2 lm1m2 m1+m2

=

langl12

l22

mL12

mL22

∣∣∣∣∣ l2

mL1 +mL2

2

ranglangl12

l22

mR12

mR22

∣∣∣∣∣ l2

mR1 +mR22






Yn12

n12n12

(x1)Yn22

n22n22



(x4) = zn11 zn2

2 zn33 zn4

4 (B22)


zn11 zn2

2 zn33 zn4


note that





that zn11 zn2

2 zn33 zn4


Yn12

n12n1minus2

2

(x1)Yn22

n2minus22

n22

(x2)Yn32


2

(x3)Yn42

minusn4minus22

minusn42

(x4)

=radicn1n2n3n4z

n1minus11 w1z

n2minus12 w2z

n3minus13 w3z

n4minus14 w4 (B24)


coefficients

Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B25)

Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B26)

ndash 32 ndash

JHEP03(2016)165


Cn12n22n1+n2

2n12n2minus2

2n1+n2minus2

2

=

radicn2

n1 + n2(B27)

Cn12n22n1+n2minus2

2n12n2minus2

2n1+n2minus2

2

= minusradic

n1n1 + n2

(B28)



n2minus12 w2z

n3minus13 w3z

n4minus14 w4 is

radicn1n2n3n4

[Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2 + 1

+Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2

]=


(B29)




section 5






parth








prime4vminus) (C1)



(2ixprime3microx

prime3 middot

part


part


)eiKmiddotx

prime3vminus

= ix23part

partxmicro3

1

x23eiKmiddotx

prime3vminus (C2)


parth


part

partxmicro2

(1

x22h

)minus x23

part

partxmicro3

(1

x23h

)minus x24

part

partxmicro4

(1

x24h

)(C3)

ndash 33 ndash

JHEP03(2016)165


x22x23x

24

(part

partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)[h

x22x23x

24

]= 0 (C4)


x1micropart

partx1νminus x1ν

part

partx1micro

)h= F1(Mmicroνe









x22x23x

24

(x1micro

part

partx1νminus x1ν

part


part

partx2νminus x2ν

part


part

partx3νminus x3ν

part

partx3micro

+x4micropart

partx4νminus x4ν

part

partx4micro

)[h

x22x23x

24

]= 0 (C6)


consider(x1 middot

part

partx1+ 3

)h = F1(e






x22x23x

24

(x1 middot

part

partx1+ x2 middot

part

partx2+ x3 middot

part

partx3+ x4 middot

part

partx4+ 12

)[h

x22x23x

24

]= 0 (C8)


h

x22x23x

24

=A




we have




ndash 34 ndash

JHEP03(2016)165





infinsumnm=0

im(minus1)n




where







minus (C14)

we find


and

infinsumn=0

(minus1)n




1


(C16)



=infinsum

n1n3n4=0

1



slot we get

α(P middot x2) =1


ndash 35 ndash

JHEP03(2016)165




=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)


=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)

(n1 + n3 + n4)



=infinsumn1

(minus1)n1




timesinfinsumn3

(minus1)n3




timesinfinsumn4

(minus1)n4





=1


1


1


=x22x

33x

24






References














ndash 36 ndash

JHEP03(2016)165










(2003) 013 [hep-th0212208] [INSPIRE]




(2014) 302 [arXiv14036646] [INSPIRE]






















(2015) 006 [arXiv150208005] [INSPIRE]






ndash 37 ndash

JHEP03(2016)165



[INSPIRE]

















(2015) 29FT01 [arXiv150500963] [INSPIRE]








[INSPIRE]














ndash 38 ndash

JHEP03(2016)165















ndash 39 ndash

Introduction























JHEP03(2016)165








1

2π2

intdS3Y l1

m1(x)Y l2

m2(x) = gm1m2

δl1l2

(l1 + 1)(A1)




gm1m2 = δm1m2


m2(x) = Y l

m1(x)

Y ml (x) = Y l

m(x) (A2)






(mLmR)

gm1m2 = δm1minusm2

Y ml (x) = Y l

minusm(x) (A3)



ables are

z = x1 + ix2 harr(

1

2

1

2


(minus1

2minus1

2

)w = x3 + ix4 harr

(1

2minus1

2


(minus1

21

2

)(A4)

We find that

Y l(l2l2) = (x1 + ix2)

l = zl (A5)

has normalization

1

2π2

intdS3(Y l

( l2 l2))lowast Y l

( l2 l2)

=1

(l + 1)(A6)


ndash 26 ndash

JHEP03(2016)165




(minusi)l

2llY lm(x)Y m

l (P )v+ (A7)


η(Y l1m1

(P )v+ Y m2l2

(K)vminus)

= δl1l2δm1m222l1(l1)2 (A8)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)= δl1l2δm1m2 (A9)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)=

1

2π2

intS3

Y m2l2

(xprime)


m1(x)) (A10)





Y lm(xprime)

xprime2


m(x) (A12)


η = f η


1


summ

1

y2Y lm(x)Y m


=infinsuml=0

summ

1

x2Y lm(xprime)Y m

l (y) if |x| gt |y| (A13)






ndash 27 ndash

JHEP03(2016)165


C2 = minus1

2LMNL


2MpqMqp (A14)






(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)eminusiP middotxv+

=

(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)ipart





Mpq = xppart

partxqminus xq

part

partxp(A16)


1

2MpqMqp(r

AY lm(x)) =

rA

2MpqMqpY


m(x) (A17)





m(x) (A19)

Thus we have

C2 rAY l




m(x) (A21)





F1 =1

λlog

((1 + ρs)


)(B1)

ndash 28 ndash

JHEP03(2016)165


of s to obtain

F1 =log(1+u)

u+

s


s2

u5(3u(2+u)minus 6 + 6u+ u2) log(1+u)

)+s3


)(B2)


F1 =

(minus1 +

u

2minus u2

3+u3


)+ s

(minus1

6+u

6minus 3u2


)+ s2

(minus1

30+

u

20minus 2u2


)+s3

(minus1

140+

u

70minus 5u2



The term at sk is

sk









in Pk(u) is

(2k)

(k)2(B6)


Qk(u) =ksuml=0

blul (B7)


ksuml=0

bl(minus1)l

(iminus l)= 0 (B8)




b1 =5b02 b2 =

20b09 b3 =

5b06 b4 =

5b042 b5 =

b0252

(B9)

ndash 29 ndash

JHEP03(2016)165



ksuml=0

MIlbl = 0 (B10)

where

MIl =(minus1)l




MIlbl (B12)



bl =sumI




bIb0 = 1 16bIb0 = 32 35 120bIb0 = 2 97 27 170

bIb0 =

5

2

20

95

6

5

42

1

252

bIb0 =

3

75

22

20

11

5

11

1

22

1

924

bIb0 =

7

2

63

13175

52

175

143

63

286

7

429

1

3432

(B14)



b1(k)b0 = k2

bk(k)b0 =k2

(2k)

bkminus1(k)b0 =k2

(2k)k(k + 1)

bkminus2(k)b0 =k2

(2k)

(k minus 1)k(k + 1)(k + 2)

4

bkminus3(k)b0 =k2

(2k)

(k

3

)(k + 3

3

)bkminusm(k)b0 =

k2

(2k)

(k

m

)(k +m

m

)(B15)

ndash 30 ndash

JHEP03(2016)165


so that

we have

b1(k) = k2(2k)

k2

bk(k) = 1



4

bkminus3(k) =

(k

3

)(k + 3

3

)bkminusm(k) =

(k

m

)(k +m

m

)(B16)


Pk(u) =

ksumi=1

aiuiminus1 (B17)



bl(minus1)iminusl

(iminus l)= 0 (B18)




k = 2 ~a = a1 ak = 6 3

k = 3 ~a =1

3times 60 60 11

k = 4 ~a =5

6times 84 126 52 5

k = 5 ~a =1

30times 7560 15120 9870 2310 137

k = 6 ~a =7

10times 13720 3300 2960 1140 174 7 (B19)

which leads to

a1(k) =(2k)

(k)2

ak(k) = 2h(k) = 2

ksuml=1

1

l(B20)




ndash 31 ndash

JHEP03(2016)165


grals






infinsumli=0

summi

infinsumt=0

sump


(x1)Yl2minusm2

(x2)Yl3m3

(xprime3)Yl4m4



t+ 1

=x23x

24

x213x224

1

λlog

((1 + ρs)


)(B21)



C l1 l2 lm1m2 m1+m2

=

langl12

l22

mL12

mL22

∣∣∣∣∣ l2

mL1 +mL2

2

ranglangl12

l22

mR12

mR22

∣∣∣∣∣ l2

mR1 +mR22






Yn12

n12n12

(x1)Yn22

n22n22



(x4) = zn11 zn2

2 zn33 zn4

4 (B22)


zn11 zn2

2 zn33 zn4


note that





that zn11 zn2

2 zn33 zn4


Yn12

n12n1minus2

2

(x1)Yn22

n2minus22

n22

(x2)Yn32


2

(x3)Yn42

minusn4minus22

minusn42

(x4)

=radicn1n2n3n4z

n1minus11 w1z

n2minus12 w2z

n3minus13 w3z

n4minus14 w4 (B24)


coefficients

Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B25)

Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B26)

ndash 32 ndash

JHEP03(2016)165


Cn12n22n1+n2

2n12n2minus2

2n1+n2minus2

2

=

radicn2

n1 + n2(B27)

Cn12n22n1+n2minus2

2n12n2minus2

2n1+n2minus2

2

= minusradic

n1n1 + n2

(B28)



n2minus12 w2z

n3minus13 w3z

n4minus14 w4 is

radicn1n2n3n4

[Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2 + 1

+Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2

]=


(B29)




section 5






parth








prime4vminus) (C1)



(2ixprime3microx

prime3 middot

part


part


)eiKmiddotx

prime3vminus

= ix23part

partxmicro3

1

x23eiKmiddotx

prime3vminus (C2)


parth


part

partxmicro2

(1

x22h

)minus x23

part

partxmicro3

(1

x23h

)minus x24

part

partxmicro4

(1

x24h

)(C3)

ndash 33 ndash

JHEP03(2016)165


x22x23x

24

(part

partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)[h

x22x23x

24

]= 0 (C4)


x1micropart

partx1νminus x1ν

part

partx1micro

)h= F1(Mmicroνe









x22x23x

24

(x1micro

part

partx1νminus x1ν

part


part

partx2νminus x2ν

part


part

partx3νminus x3ν

part

partx3micro

+x4micropart

partx4νminus x4ν

part

partx4micro

)[h

x22x23x

24

]= 0 (C6)


consider(x1 middot

part

partx1+ 3

)h = F1(e






x22x23x

24

(x1 middot

part

partx1+ x2 middot

part

partx2+ x3 middot

part

partx3+ x4 middot

part

partx4+ 12

)[h

x22x23x

24

]= 0 (C8)


h

x22x23x

24

=A




we have




ndash 34 ndash

JHEP03(2016)165





infinsumnm=0

im(minus1)n




where







minus (C14)

we find


and

infinsumn=0

(minus1)n




1


(C16)



=infinsum

n1n3n4=0

1



slot we get

α(P middot x2) =1


ndash 35 ndash

JHEP03(2016)165




=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)


=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)

(n1 + n3 + n4)



=infinsumn1

(minus1)n1




timesinfinsumn3

(minus1)n3




timesinfinsumn4

(minus1)n4





=1


1


1


=x22x

33x

24






References














ndash 36 ndash

JHEP03(2016)165










(2003) 013 [hep-th0212208] [INSPIRE]




(2014) 302 [arXiv14036646] [INSPIRE]






















(2015) 006 [arXiv150208005] [INSPIRE]






ndash 37 ndash

JHEP03(2016)165



[INSPIRE]

















(2015) 29FT01 [arXiv150500963] [INSPIRE]








[INSPIRE]














ndash 38 ndash

JHEP03(2016)165















ndash 39 ndash

Introduction























JHEP03(2016)165




(minusi)l

2llY lm(x)Y m

l (P )v+ (A7)


η(Y l1m1

(P )v+ Y m2l2

(K)vminus)

= δl1l2δm1m222l1(l1)2 (A8)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)= δl1l2δm1m2 (A9)


η

(Y l1m1

(x)Y m2l2

(xprime)

xprime2

)=

1

2π2

intS3

Y m2l2

(xprime)


m1(x)) (A10)





Y lm(xprime)

xprime2


m(x) (A12)


η = f η


1


summ

1

y2Y lm(x)Y m


=infinsuml=0

summ

1

x2Y lm(xprime)Y m

l (y) if |x| gt |y| (A13)






ndash 27 ndash

JHEP03(2016)165


C2 = minus1

2LMNL


2MpqMqp (A14)






(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)eminusiP middotxv+

=

(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)ipart





Mpq = xppart

partxqminus xq

part

partxp(A16)


1

2MpqMqp(r

AY lm(x)) =

rA

2MpqMqpY


m(x) (A17)





m(x) (A19)

Thus we have

C2 rAY l




m(x) (A21)





F1 =1

λlog

((1 + ρs)


)(B1)

ndash 28 ndash

JHEP03(2016)165


of s to obtain

F1 =log(1+u)

u+

s


s2

u5(3u(2+u)minus 6 + 6u+ u2) log(1+u)

)+s3


)(B2)


F1 =

(minus1 +

u

2minus u2

3+u3


)+ s

(minus1

6+u

6minus 3u2


)+ s2

(minus1

30+

u

20minus 2u2


)+s3

(minus1

140+

u

70minus 5u2



The term at sk is

sk









in Pk(u) is

(2k)

(k)2(B6)


Qk(u) =ksuml=0

blul (B7)


ksuml=0

bl(minus1)l

(iminus l)= 0 (B8)




b1 =5b02 b2 =

20b09 b3 =

5b06 b4 =

5b042 b5 =

b0252

(B9)

ndash 29 ndash

JHEP03(2016)165



ksuml=0

MIlbl = 0 (B10)

where

MIl =(minus1)l




MIlbl (B12)



bl =sumI




bIb0 = 1 16bIb0 = 32 35 120bIb0 = 2 97 27 170

bIb0 =

5

2

20

95

6

5

42

1

252

bIb0 =

3

75

22

20

11

5

11

1

22

1

924

bIb0 =

7

2

63

13175

52

175

143

63

286

7

429

1

3432

(B14)



b1(k)b0 = k2

bk(k)b0 =k2

(2k)

bkminus1(k)b0 =k2

(2k)k(k + 1)

bkminus2(k)b0 =k2

(2k)

(k minus 1)k(k + 1)(k + 2)

4

bkminus3(k)b0 =k2

(2k)

(k

3

)(k + 3

3

)bkminusm(k)b0 =

k2

(2k)

(k

m

)(k +m

m

)(B15)

ndash 30 ndash

JHEP03(2016)165


so that

we have

b1(k) = k2(2k)

k2

bk(k) = 1



4

bkminus3(k) =

(k

3

)(k + 3

3

)bkminusm(k) =

(k

m

)(k +m

m

)(B16)


Pk(u) =

ksumi=1

aiuiminus1 (B17)



bl(minus1)iminusl

(iminus l)= 0 (B18)




k = 2 ~a = a1 ak = 6 3

k = 3 ~a =1

3times 60 60 11

k = 4 ~a =5

6times 84 126 52 5

k = 5 ~a =1

30times 7560 15120 9870 2310 137

k = 6 ~a =7

10times 13720 3300 2960 1140 174 7 (B19)

which leads to

a1(k) =(2k)

(k)2

ak(k) = 2h(k) = 2

ksuml=1

1

l(B20)




ndash 31 ndash

JHEP03(2016)165


grals






infinsumli=0

summi

infinsumt=0

sump


(x1)Yl2minusm2

(x2)Yl3m3

(xprime3)Yl4m4



t+ 1

=x23x

24

x213x224

1

λlog

((1 + ρs)


)(B21)



C l1 l2 lm1m2 m1+m2

=

langl12

l22

mL12

mL22

∣∣∣∣∣ l2

mL1 +mL2

2

ranglangl12

l22

mR12

mR22

∣∣∣∣∣ l2

mR1 +mR22






Yn12

n12n12

(x1)Yn22

n22n22



(x4) = zn11 zn2

2 zn33 zn4

4 (B22)


zn11 zn2

2 zn33 zn4


note that





that zn11 zn2

2 zn33 zn4


Yn12

n12n1minus2

2

(x1)Yn22

n2minus22

n22

(x2)Yn32


2

(x3)Yn42

minusn4minus22

minusn42

(x4)

=radicn1n2n3n4z

n1minus11 w1z

n2minus12 w2z

n3minus13 w3z

n4minus14 w4 (B24)


coefficients

Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B25)

Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B26)

ndash 32 ndash

JHEP03(2016)165


Cn12n22n1+n2

2n12n2minus2

2n1+n2minus2

2

=

radicn2

n1 + n2(B27)

Cn12n22n1+n2minus2

2n12n2minus2

2n1+n2minus2

2

= minusradic

n1n1 + n2

(B28)



n2minus12 w2z

n3minus13 w3z

n4minus14 w4 is

radicn1n2n3n4

[Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2 + 1

+Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2

]=


(B29)




section 5






parth








prime4vminus) (C1)



(2ixprime3microx

prime3 middot

part


part


)eiKmiddotx

prime3vminus

= ix23part

partxmicro3

1

x23eiKmiddotx

prime3vminus (C2)


parth


part

partxmicro2

(1

x22h

)minus x23

part

partxmicro3

(1

x23h

)minus x24

part

partxmicro4

(1

x24h

)(C3)

ndash 33 ndash

JHEP03(2016)165


x22x23x

24

(part

partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)[h

x22x23x

24

]= 0 (C4)


x1micropart

partx1νminus x1ν

part

partx1micro

)h= F1(Mmicroνe









x22x23x

24

(x1micro

part

partx1νminus x1ν

part


part

partx2νminus x2ν

part


part

partx3νminus x3ν

part

partx3micro

+x4micropart

partx4νminus x4ν

part

partx4micro

)[h

x22x23x

24

]= 0 (C6)


consider(x1 middot

part

partx1+ 3

)h = F1(e






x22x23x

24

(x1 middot

part

partx1+ x2 middot

part

partx2+ x3 middot

part

partx3+ x4 middot

part

partx4+ 12

)[h

x22x23x

24

]= 0 (C8)


h

x22x23x

24

=A




we have




ndash 34 ndash

JHEP03(2016)165





infinsumnm=0

im(minus1)n




where







minus (C14)

we find


and

infinsumn=0

(minus1)n




1


(C16)



=infinsum

n1n3n4=0

1



slot we get

α(P middot x2) =1


ndash 35 ndash

JHEP03(2016)165




=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)


=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)

(n1 + n3 + n4)



=infinsumn1

(minus1)n1




timesinfinsumn3

(minus1)n3




timesinfinsumn4

(minus1)n4





=1


1


1


=x22x

33x

24






References














ndash 36 ndash

JHEP03(2016)165










(2003) 013 [hep-th0212208] [INSPIRE]




(2014) 302 [arXiv14036646] [INSPIRE]






















(2015) 006 [arXiv150208005] [INSPIRE]






ndash 37 ndash

JHEP03(2016)165



[INSPIRE]

















(2015) 29FT01 [arXiv150500963] [INSPIRE]








[INSPIRE]














ndash 38 ndash

JHEP03(2016)165















ndash 39 ndash

Introduction























JHEP03(2016)165


C2 = minus1

2LMNL


2MpqMqp (A14)






(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)eminusiP middotxv+

=

(2ixpx middot

part

partxminus ix2 part

partxp+ 2ixp

)ipart





Mpq = xppart

partxqminus xq

part

partxp(A16)


1

2MpqMqp(r

AY lm(x)) =

rA

2MpqMqpY


m(x) (A17)





m(x) (A19)

Thus we have

C2 rAY l




m(x) (A21)





F1 =1

λlog

((1 + ρs)


)(B1)

ndash 28 ndash

JHEP03(2016)165


of s to obtain

F1 =log(1+u)

u+

s


s2

u5(3u(2+u)minus 6 + 6u+ u2) log(1+u)

)+s3


)(B2)


F1 =

(minus1 +

u

2minus u2

3+u3


)+ s

(minus1

6+u

6minus 3u2


)+ s2

(minus1

30+

u

20minus 2u2


)+s3

(minus1

140+

u

70minus 5u2



The term at sk is

sk









in Pk(u) is

(2k)

(k)2(B6)


Qk(u) =ksuml=0

blul (B7)


ksuml=0

bl(minus1)l

(iminus l)= 0 (B8)




b1 =5b02 b2 =

20b09 b3 =

5b06 b4 =

5b042 b5 =

b0252

(B9)

ndash 29 ndash

JHEP03(2016)165



ksuml=0

MIlbl = 0 (B10)

where

MIl =(minus1)l




MIlbl (B12)



bl =sumI




bIb0 = 1 16bIb0 = 32 35 120bIb0 = 2 97 27 170

bIb0 =

5

2

20

95

6

5

42

1

252

bIb0 =

3

75

22

20

11

5

11

1

22

1

924

bIb0 =

7

2

63

13175

52

175

143

63

286

7

429

1

3432

(B14)



b1(k)b0 = k2

bk(k)b0 =k2

(2k)

bkminus1(k)b0 =k2

(2k)k(k + 1)

bkminus2(k)b0 =k2

(2k)

(k minus 1)k(k + 1)(k + 2)

4

bkminus3(k)b0 =k2

(2k)

(k

3

)(k + 3

3

)bkminusm(k)b0 =

k2

(2k)

(k

m

)(k +m

m

)(B15)

ndash 30 ndash

JHEP03(2016)165


so that

we have

b1(k) = k2(2k)

k2

bk(k) = 1



4

bkminus3(k) =

(k

3

)(k + 3

3

)bkminusm(k) =

(k

m

)(k +m

m

)(B16)


Pk(u) =

ksumi=1

aiuiminus1 (B17)



bl(minus1)iminusl

(iminus l)= 0 (B18)




k = 2 ~a = a1 ak = 6 3

k = 3 ~a =1

3times 60 60 11

k = 4 ~a =5

6times 84 126 52 5

k = 5 ~a =1

30times 7560 15120 9870 2310 137

k = 6 ~a =7

10times 13720 3300 2960 1140 174 7 (B19)

which leads to

a1(k) =(2k)

(k)2

ak(k) = 2h(k) = 2

ksuml=1

1

l(B20)




ndash 31 ndash

JHEP03(2016)165


grals






infinsumli=0

summi

infinsumt=0

sump


(x1)Yl2minusm2

(x2)Yl3m3

(xprime3)Yl4m4



t+ 1

=x23x

24

x213x224

1

λlog

((1 + ρs)


)(B21)



C l1 l2 lm1m2 m1+m2

=

langl12

l22

mL12

mL22

∣∣∣∣∣ l2

mL1 +mL2

2

ranglangl12

l22

mR12

mR22

∣∣∣∣∣ l2

mR1 +mR22






Yn12

n12n12

(x1)Yn22

n22n22



(x4) = zn11 zn2

2 zn33 zn4

4 (B22)


zn11 zn2

2 zn33 zn4


note that





that zn11 zn2

2 zn33 zn4


Yn12

n12n1minus2

2

(x1)Yn22

n2minus22

n22

(x2)Yn32


2

(x3)Yn42

minusn4minus22

minusn42

(x4)

=radicn1n2n3n4z

n1minus11 w1z

n2minus12 w2z

n3minus13 w3z

n4minus14 w4 (B24)


coefficients

Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B25)

Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B26)

ndash 32 ndash

JHEP03(2016)165


Cn12n22n1+n2

2n12n2minus2

2n1+n2minus2

2

=

radicn2

n1 + n2(B27)

Cn12n22n1+n2minus2

2n12n2minus2

2n1+n2minus2

2

= minusradic

n1n1 + n2

(B28)



n2minus12 w2z

n3minus13 w3z

n4minus14 w4 is

radicn1n2n3n4

[Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2 + 1

+Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2

]=


(B29)




section 5






parth








prime4vminus) (C1)



(2ixprime3microx

prime3 middot

part


part


)eiKmiddotx

prime3vminus

= ix23part

partxmicro3

1

x23eiKmiddotx

prime3vminus (C2)


parth


part

partxmicro2

(1

x22h

)minus x23

part

partxmicro3

(1

x23h

)minus x24

part

partxmicro4

(1

x24h

)(C3)

ndash 33 ndash

JHEP03(2016)165


x22x23x

24

(part

partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)[h

x22x23x

24

]= 0 (C4)


x1micropart

partx1νminus x1ν

part

partx1micro

)h= F1(Mmicroνe









x22x23x

24

(x1micro

part

partx1νminus x1ν

part


part

partx2νminus x2ν

part


part

partx3νminus x3ν

part

partx3micro

+x4micropart

partx4νminus x4ν

part

partx4micro

)[h

x22x23x

24

]= 0 (C6)


consider(x1 middot

part

partx1+ 3

)h = F1(e






x22x23x

24

(x1 middot

part

partx1+ x2 middot

part

partx2+ x3 middot

part

partx3+ x4 middot

part

partx4+ 12

)[h

x22x23x

24

]= 0 (C8)


h

x22x23x

24

=A




we have




ndash 34 ndash

JHEP03(2016)165





infinsumnm=0

im(minus1)n




where







minus (C14)

we find


and

infinsumn=0

(minus1)n




1


(C16)



=infinsum

n1n3n4=0

1



slot we get

α(P middot x2) =1


ndash 35 ndash

JHEP03(2016)165




=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)


=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)

(n1 + n3 + n4)



=infinsumn1

(minus1)n1




timesinfinsumn3

(minus1)n3




timesinfinsumn4

(minus1)n4





=1


1


1


=x22x

33x

24






References














ndash 36 ndash

JHEP03(2016)165










(2003) 013 [hep-th0212208] [INSPIRE]




(2014) 302 [arXiv14036646] [INSPIRE]






















(2015) 006 [arXiv150208005] [INSPIRE]






ndash 37 ndash

JHEP03(2016)165



[INSPIRE]

















(2015) 29FT01 [arXiv150500963] [INSPIRE]








[INSPIRE]














ndash 38 ndash

JHEP03(2016)165















ndash 39 ndash

Introduction























JHEP03(2016)165


of s to obtain

F1 =log(1+u)

u+

s


s2

u5(3u(2+u)minus 6 + 6u+ u2) log(1+u)

)+s3


)(B2)


F1 =

(minus1 +

u

2minus u2

3+u3


)+ s

(minus1

6+u

6minus 3u2


)+ s2

(minus1

30+

u

20minus 2u2


)+s3

(minus1

140+

u

70minus 5u2



The term at sk is

sk









in Pk(u) is

(2k)

(k)2(B6)


Qk(u) =ksuml=0

blul (B7)


ksuml=0

bl(minus1)l

(iminus l)= 0 (B8)




b1 =5b02 b2 =

20b09 b3 =

5b06 b4 =

5b042 b5 =

b0252

(B9)

ndash 29 ndash

JHEP03(2016)165



ksuml=0

MIlbl = 0 (B10)

where

MIl =(minus1)l




MIlbl (B12)



bl =sumI




bIb0 = 1 16bIb0 = 32 35 120bIb0 = 2 97 27 170

bIb0 =

5

2

20

95

6

5

42

1

252

bIb0 =

3

75

22

20

11

5

11

1

22

1

924

bIb0 =

7

2

63

13175

52

175

143

63

286

7

429

1

3432

(B14)



b1(k)b0 = k2

bk(k)b0 =k2

(2k)

bkminus1(k)b0 =k2

(2k)k(k + 1)

bkminus2(k)b0 =k2

(2k)

(k minus 1)k(k + 1)(k + 2)

4

bkminus3(k)b0 =k2

(2k)

(k

3

)(k + 3

3

)bkminusm(k)b0 =

k2

(2k)

(k

m

)(k +m

m

)(B15)

ndash 30 ndash

JHEP03(2016)165


so that

we have

b1(k) = k2(2k)

k2

bk(k) = 1



4

bkminus3(k) =

(k

3

)(k + 3

3

)bkminusm(k) =

(k

m

)(k +m

m

)(B16)


Pk(u) =

ksumi=1

aiuiminus1 (B17)



bl(minus1)iminusl

(iminus l)= 0 (B18)




k = 2 ~a = a1 ak = 6 3

k = 3 ~a =1

3times 60 60 11

k = 4 ~a =5

6times 84 126 52 5

k = 5 ~a =1

30times 7560 15120 9870 2310 137

k = 6 ~a =7

10times 13720 3300 2960 1140 174 7 (B19)

which leads to

a1(k) =(2k)

(k)2

ak(k) = 2h(k) = 2

ksuml=1

1

l(B20)




ndash 31 ndash

JHEP03(2016)165


grals






infinsumli=0

summi

infinsumt=0

sump


(x1)Yl2minusm2

(x2)Yl3m3

(xprime3)Yl4m4



t+ 1

=x23x

24

x213x224

1

λlog

((1 + ρs)


)(B21)



C l1 l2 lm1m2 m1+m2

=

langl12

l22

mL12

mL22

∣∣∣∣∣ l2

mL1 +mL2

2

ranglangl12

l22

mR12

mR22

∣∣∣∣∣ l2

mR1 +mR22






Yn12

n12n12

(x1)Yn22

n22n22



(x4) = zn11 zn2

2 zn33 zn4

4 (B22)


zn11 zn2

2 zn33 zn4


note that





that zn11 zn2

2 zn33 zn4


Yn12

n12n1minus2

2

(x1)Yn22

n2minus22

n22

(x2)Yn32


2

(x3)Yn42

minusn4minus22

minusn42

(x4)

=radicn1n2n3n4z

n1minus11 w1z

n2minus12 w2z

n3minus13 w3z

n4minus14 w4 (B24)


coefficients

Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B25)

Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B26)

ndash 32 ndash

JHEP03(2016)165


Cn12n22n1+n2

2n12n2minus2

2n1+n2minus2

2

=

radicn2

n1 + n2(B27)

Cn12n22n1+n2minus2

2n12n2minus2

2n1+n2minus2

2

= minusradic

n1n1 + n2

(B28)



n2minus12 w2z

n3minus13 w3z

n4minus14 w4 is

radicn1n2n3n4

[Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2 + 1

+Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2

]=


(B29)




section 5






parth








prime4vminus) (C1)



(2ixprime3microx

prime3 middot

part


part


)eiKmiddotx

prime3vminus

= ix23part

partxmicro3

1

x23eiKmiddotx

prime3vminus (C2)


parth


part

partxmicro2

(1

x22h

)minus x23

part

partxmicro3

(1

x23h

)minus x24

part

partxmicro4

(1

x24h

)(C3)

ndash 33 ndash

JHEP03(2016)165


x22x23x

24

(part

partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)[h

x22x23x

24

]= 0 (C4)


x1micropart

partx1νminus x1ν

part

partx1micro

)h= F1(Mmicroνe









x22x23x

24

(x1micro

part

partx1νminus x1ν

part


part

partx2νminus x2ν

part


part

partx3νminus x3ν

part

partx3micro

+x4micropart

partx4νminus x4ν

part

partx4micro

)[h

x22x23x

24

]= 0 (C6)


consider(x1 middot

part

partx1+ 3

)h = F1(e






x22x23x

24

(x1 middot

part

partx1+ x2 middot

part

partx2+ x3 middot

part

partx3+ x4 middot

part

partx4+ 12

)[h

x22x23x

24

]= 0 (C8)


h

x22x23x

24

=A




we have




ndash 34 ndash

JHEP03(2016)165





infinsumnm=0

im(minus1)n




where







minus (C14)

we find


and

infinsumn=0

(minus1)n




1


(C16)



=infinsum

n1n3n4=0

1



slot we get

α(P middot x2) =1


ndash 35 ndash

JHEP03(2016)165




=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)


=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)

(n1 + n3 + n4)



=infinsumn1

(minus1)n1




timesinfinsumn3

(minus1)n3




timesinfinsumn4

(minus1)n4





=1


1


1


=x22x

33x

24






References














ndash 36 ndash

JHEP03(2016)165










(2003) 013 [hep-th0212208] [INSPIRE]




(2014) 302 [arXiv14036646] [INSPIRE]






















(2015) 006 [arXiv150208005] [INSPIRE]






ndash 37 ndash

JHEP03(2016)165



[INSPIRE]

















(2015) 29FT01 [arXiv150500963] [INSPIRE]








[INSPIRE]














ndash 38 ndash

JHEP03(2016)165















ndash 39 ndash

Introduction























JHEP03(2016)165



ksuml=0

MIlbl = 0 (B10)

where

MIl =(minus1)l




MIlbl (B12)



bl =sumI




bIb0 = 1 16bIb0 = 32 35 120bIb0 = 2 97 27 170

bIb0 =

5

2

20

95

6

5

42

1

252

bIb0 =

3

75

22

20

11

5

11

1

22

1

924

bIb0 =

7

2

63

13175

52

175

143

63

286

7

429

1

3432

(B14)



b1(k)b0 = k2

bk(k)b0 =k2

(2k)

bkminus1(k)b0 =k2

(2k)k(k + 1)

bkminus2(k)b0 =k2

(2k)

(k minus 1)k(k + 1)(k + 2)

4

bkminus3(k)b0 =k2

(2k)

(k

3

)(k + 3

3

)bkminusm(k)b0 =

k2

(2k)

(k

m

)(k +m

m

)(B15)

ndash 30 ndash

JHEP03(2016)165


so that

we have

b1(k) = k2(2k)

k2

bk(k) = 1



4

bkminus3(k) =

(k

3

)(k + 3

3

)bkminusm(k) =

(k

m

)(k +m

m

)(B16)


Pk(u) =

ksumi=1

aiuiminus1 (B17)



bl(minus1)iminusl

(iminus l)= 0 (B18)




k = 2 ~a = a1 ak = 6 3

k = 3 ~a =1

3times 60 60 11

k = 4 ~a =5

6times 84 126 52 5

k = 5 ~a =1

30times 7560 15120 9870 2310 137

k = 6 ~a =7

10times 13720 3300 2960 1140 174 7 (B19)

which leads to

a1(k) =(2k)

(k)2

ak(k) = 2h(k) = 2

ksuml=1

1

l(B20)




ndash 31 ndash

JHEP03(2016)165


grals






infinsumli=0

summi

infinsumt=0

sump


(x1)Yl2minusm2

(x2)Yl3m3

(xprime3)Yl4m4



t+ 1

=x23x

24

x213x224

1

λlog

((1 + ρs)


)(B21)



C l1 l2 lm1m2 m1+m2

=

langl12

l22

mL12

mL22

∣∣∣∣∣ l2

mL1 +mL2

2

ranglangl12

l22

mR12

mR22

∣∣∣∣∣ l2

mR1 +mR22






Yn12

n12n12

(x1)Yn22

n22n22



(x4) = zn11 zn2

2 zn33 zn4

4 (B22)


zn11 zn2

2 zn33 zn4


note that





that zn11 zn2

2 zn33 zn4


Yn12

n12n1minus2

2

(x1)Yn22

n2minus22

n22

(x2)Yn32


2

(x3)Yn42

minusn4minus22

minusn42

(x4)

=radicn1n2n3n4z

n1minus11 w1z

n2minus12 w2z

n3minus13 w3z

n4minus14 w4 (B24)


coefficients

Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B25)

Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B26)

ndash 32 ndash

JHEP03(2016)165


Cn12n22n1+n2

2n12n2minus2

2n1+n2minus2

2

=

radicn2

n1 + n2(B27)

Cn12n22n1+n2minus2

2n12n2minus2

2n1+n2minus2

2

= minusradic

n1n1 + n2

(B28)



n2minus12 w2z

n3minus13 w3z

n4minus14 w4 is

radicn1n2n3n4

[Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2 + 1

+Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2

]=


(B29)




section 5






parth








prime4vminus) (C1)



(2ixprime3microx

prime3 middot

part


part


)eiKmiddotx

prime3vminus

= ix23part

partxmicro3

1

x23eiKmiddotx

prime3vminus (C2)


parth


part

partxmicro2

(1

x22h

)minus x23

part

partxmicro3

(1

x23h

)minus x24

part

partxmicro4

(1

x24h

)(C3)

ndash 33 ndash

JHEP03(2016)165


x22x23x

24

(part

partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)[h

x22x23x

24

]= 0 (C4)


x1micropart

partx1νminus x1ν

part

partx1micro

)h= F1(Mmicroνe









x22x23x

24

(x1micro

part

partx1νminus x1ν

part


part

partx2νminus x2ν

part


part

partx3νminus x3ν

part

partx3micro

+x4micropart

partx4νminus x4ν

part

partx4micro

)[h

x22x23x

24

]= 0 (C6)


consider(x1 middot

part

partx1+ 3

)h = F1(e






x22x23x

24

(x1 middot

part

partx1+ x2 middot

part

partx2+ x3 middot

part

partx3+ x4 middot

part

partx4+ 12

)[h

x22x23x

24

]= 0 (C8)


h

x22x23x

24

=A




we have




ndash 34 ndash

JHEP03(2016)165





infinsumnm=0

im(minus1)n




where







minus (C14)

we find


and

infinsumn=0

(minus1)n




1


(C16)



=infinsum

n1n3n4=0

1



slot we get

α(P middot x2) =1


ndash 35 ndash

JHEP03(2016)165




=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)


=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)

(n1 + n3 + n4)



=infinsumn1

(minus1)n1




timesinfinsumn3

(minus1)n3




timesinfinsumn4

(minus1)n4





=1


1


1


=x22x

33x

24






References














ndash 36 ndash

JHEP03(2016)165










(2003) 013 [hep-th0212208] [INSPIRE]




(2014) 302 [arXiv14036646] [INSPIRE]






















(2015) 006 [arXiv150208005] [INSPIRE]






ndash 37 ndash

JHEP03(2016)165



[INSPIRE]

















(2015) 29FT01 [arXiv150500963] [INSPIRE]








[INSPIRE]














ndash 38 ndash

JHEP03(2016)165















ndash 39 ndash

Introduction























JHEP03(2016)165


so that

we have

b1(k) = k2(2k)

k2

bk(k) = 1



4

bkminus3(k) =

(k

3

)(k + 3

3

)bkminusm(k) =

(k

m

)(k +m

m

)(B16)


Pk(u) =

ksumi=1

aiuiminus1 (B17)



bl(minus1)iminusl

(iminus l)= 0 (B18)




k = 2 ~a = a1 ak = 6 3

k = 3 ~a =1

3times 60 60 11

k = 4 ~a =5

6times 84 126 52 5

k = 5 ~a =1

30times 7560 15120 9870 2310 137

k = 6 ~a =7

10times 13720 3300 2960 1140 174 7 (B19)

which leads to

a1(k) =(2k)

(k)2

ak(k) = 2h(k) = 2

ksuml=1

1

l(B20)




ndash 31 ndash

JHEP03(2016)165


grals






infinsumli=0

summi

infinsumt=0

sump


(x1)Yl2minusm2

(x2)Yl3m3

(xprime3)Yl4m4



t+ 1

=x23x

24

x213x224

1

λlog

((1 + ρs)


)(B21)



C l1 l2 lm1m2 m1+m2

=

langl12

l22

mL12

mL22

∣∣∣∣∣ l2

mL1 +mL2

2

ranglangl12

l22

mR12

mR22

∣∣∣∣∣ l2

mR1 +mR22






Yn12

n12n12

(x1)Yn22

n22n22



(x4) = zn11 zn2

2 zn33 zn4

4 (B22)


zn11 zn2

2 zn33 zn4


note that





that zn11 zn2

2 zn33 zn4


Yn12

n12n1minus2

2

(x1)Yn22

n2minus22

n22

(x2)Yn32


2

(x3)Yn42

minusn4minus22

minusn42

(x4)

=radicn1n2n3n4z

n1minus11 w1z

n2minus12 w2z

n3minus13 w3z

n4minus14 w4 (B24)


coefficients

Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B25)

Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B26)

ndash 32 ndash

JHEP03(2016)165


Cn12n22n1+n2

2n12n2minus2

2n1+n2minus2

2

=

radicn2

n1 + n2(B27)

Cn12n22n1+n2minus2

2n12n2minus2

2n1+n2minus2

2

= minusradic

n1n1 + n2

(B28)



n2minus12 w2z

n3minus13 w3z

n4minus14 w4 is

radicn1n2n3n4

[Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2 + 1

+Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2

]=


(B29)




section 5






parth








prime4vminus) (C1)



(2ixprime3microx

prime3 middot

part


part


)eiKmiddotx

prime3vminus

= ix23part

partxmicro3

1

x23eiKmiddotx

prime3vminus (C2)


parth


part

partxmicro2

(1

x22h

)minus x23

part

partxmicro3

(1

x23h

)minus x24

part

partxmicro4

(1

x24h

)(C3)

ndash 33 ndash

JHEP03(2016)165


x22x23x

24

(part

partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)[h

x22x23x

24

]= 0 (C4)


x1micropart

partx1νminus x1ν

part

partx1micro

)h= F1(Mmicroνe









x22x23x

24

(x1micro

part

partx1νminus x1ν

part


part

partx2νminus x2ν

part


part

partx3νminus x3ν

part

partx3micro

+x4micropart

partx4νminus x4ν

part

partx4micro

)[h

x22x23x

24

]= 0 (C6)


consider(x1 middot

part

partx1+ 3

)h = F1(e






x22x23x

24

(x1 middot

part

partx1+ x2 middot

part

partx2+ x3 middot

part

partx3+ x4 middot

part

partx4+ 12

)[h

x22x23x

24

]= 0 (C8)


h

x22x23x

24

=A




we have




ndash 34 ndash

JHEP03(2016)165





infinsumnm=0

im(minus1)n




where







minus (C14)

we find


and

infinsumn=0

(minus1)n




1


(C16)



=infinsum

n1n3n4=0

1



slot we get

α(P middot x2) =1


ndash 35 ndash

JHEP03(2016)165




=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)


=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)

(n1 + n3 + n4)



=infinsumn1

(minus1)n1




timesinfinsumn3

(minus1)n3




timesinfinsumn4

(minus1)n4





=1


1


1


=x22x

33x

24






References














ndash 36 ndash

JHEP03(2016)165










(2003) 013 [hep-th0212208] [INSPIRE]




(2014) 302 [arXiv14036646] [INSPIRE]






















(2015) 006 [arXiv150208005] [INSPIRE]






ndash 37 ndash

JHEP03(2016)165



[INSPIRE]

















(2015) 29FT01 [arXiv150500963] [INSPIRE]








[INSPIRE]














ndash 38 ndash

JHEP03(2016)165















ndash 39 ndash

Introduction























JHEP03(2016)165


grals






infinsumli=0

summi

infinsumt=0

sump


(x1)Yl2minusm2

(x2)Yl3m3

(xprime3)Yl4m4



t+ 1

=x23x

24

x213x224

1

λlog

((1 + ρs)


)(B21)



C l1 l2 lm1m2 m1+m2

=

langl12

l22

mL12

mL22

∣∣∣∣∣ l2

mL1 +mL2

2

ranglangl12

l22

mR12

mR22

∣∣∣∣∣ l2

mR1 +mR22






Yn12

n12n12

(x1)Yn22

n22n22



(x4) = zn11 zn2

2 zn33 zn4

4 (B22)


zn11 zn2

2 zn33 zn4


note that





that zn11 zn2

2 zn33 zn4


Yn12

n12n1minus2

2

(x1)Yn22

n2minus22

n22

(x2)Yn32


2

(x3)Yn42

minusn4minus22

minusn42

(x4)

=radicn1n2n3n4z

n1minus11 w1z

n2minus12 w2z

n3minus13 w3z

n4minus14 w4 (B24)


coefficients

Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B25)

Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) =

radicn1n2

n1 + n2(B26)

ndash 32 ndash

JHEP03(2016)165


Cn12n22n1+n2

2n12n2minus2

2n1+n2minus2

2

=

radicn2

n1 + n2(B27)

Cn12n22n1+n2minus2

2n12n2minus2

2n1+n2minus2

2

= minusradic

n1n1 + n2

(B28)



n2minus12 w2z

n3minus13 w3z

n4minus14 w4 is

radicn1n2n3n4

[Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2 + 1

+Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2

]=


(B29)




section 5






parth








prime4vminus) (C1)



(2ixprime3microx

prime3 middot

part


part


)eiKmiddotx

prime3vminus

= ix23part

partxmicro3

1

x23eiKmiddotx

prime3vminus (C2)


parth


part

partxmicro2

(1

x22h

)minus x23

part

partxmicro3

(1

x23h

)minus x24

part

partxmicro4

(1

x24h

)(C3)

ndash 33 ndash

JHEP03(2016)165


x22x23x

24

(part

partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)[h

x22x23x

24

]= 0 (C4)


x1micropart

partx1νminus x1ν

part

partx1micro

)h= F1(Mmicroνe









x22x23x

24

(x1micro

part

partx1νminus x1ν

part


part

partx2νminus x2ν

part


part

partx3νminus x3ν

part

partx3micro

+x4micropart

partx4νminus x4ν

part

partx4micro

)[h

x22x23x

24

]= 0 (C6)


consider(x1 middot

part

partx1+ 3

)h = F1(e






x22x23x

24

(x1 middot

part

partx1+ x2 middot

part

partx2+ x3 middot

part

partx3+ x4 middot

part

partx4+ 12

)[h

x22x23x

24

]= 0 (C8)


h

x22x23x

24

=A




we have




ndash 34 ndash

JHEP03(2016)165





infinsumnm=0

im(minus1)n




where







minus (C14)

we find


and

infinsumn=0

(minus1)n




1


(C16)



=infinsum

n1n3n4=0

1



slot we get

α(P middot x2) =1


ndash 35 ndash

JHEP03(2016)165




=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)


=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)

(n1 + n3 + n4)



=infinsumn1

(minus1)n1




timesinfinsumn3

(minus1)n3




timesinfinsumn4

(minus1)n4





=1


1


1


=x22x

33x

24






References














ndash 36 ndash

JHEP03(2016)165










(2003) 013 [hep-th0212208] [INSPIRE]




(2014) 302 [arXiv14036646] [INSPIRE]






















(2015) 006 [arXiv150208005] [INSPIRE]






ndash 37 ndash

JHEP03(2016)165



[INSPIRE]

















(2015) 29FT01 [arXiv150500963] [INSPIRE]








[INSPIRE]














ndash 38 ndash

JHEP03(2016)165















ndash 39 ndash

Introduction























JHEP03(2016)165


Cn12n22n1+n2

2n12n2minus2

2n1+n2minus2

2

=

radicn2

n1 + n2(B27)

Cn12n22n1+n2minus2

2n12n2minus2

2n1+n2minus2

2

= minusradic

n1n1 + n2

(B28)



n2minus12 w2z

n3minus13 w3z

n4minus14 w4 is

radicn1n2n3n4

[Cn12n22n1+n2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2 + 1

+Cn12n22n1+n2minus2

2(n12n1minus2

2

)(n2minus2

2n22

)(n1+n2minus2

2n1+n2minus2

2

) 1

n1 + n2

]=


(B29)




section 5






parth








prime4vminus) (C1)



(2ixprime3microx

prime3 middot

part


part


)eiKmiddotx

prime3vminus

= ix23part

partxmicro3

1

x23eiKmiddotx

prime3vminus (C2)


parth


part

partxmicro2

(1

x22h

)minus x23

part

partxmicro3

(1

x23h

)minus x24

part

partxmicro4

(1

x24h

)(C3)

ndash 33 ndash

JHEP03(2016)165


x22x23x

24

(part

partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)[h

x22x23x

24

]= 0 (C4)


x1micropart

partx1νminus x1ν

part

partx1micro

)h= F1(Mmicroνe









x22x23x

24

(x1micro

part

partx1νminus x1ν

part


part

partx2νminus x2ν

part


part

partx3νminus x3ν

part

partx3micro

+x4micropart

partx4νminus x4ν

part

partx4micro

)[h

x22x23x

24

]= 0 (C6)


consider(x1 middot

part

partx1+ 3

)h = F1(e






x22x23x

24

(x1 middot

part

partx1+ x2 middot

part

partx2+ x3 middot

part

partx3+ x4 middot

part

partx4+ 12

)[h

x22x23x

24

]= 0 (C8)


h

x22x23x

24

=A




we have




ndash 34 ndash

JHEP03(2016)165





infinsumnm=0

im(minus1)n




where







minus (C14)

we find


and

infinsumn=0

(minus1)n




1


(C16)



=infinsum

n1n3n4=0

1



slot we get

α(P middot x2) =1


ndash 35 ndash

JHEP03(2016)165




=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)


=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)

(n1 + n3 + n4)



=infinsumn1

(minus1)n1




timesinfinsumn3

(minus1)n3




timesinfinsumn4

(minus1)n4





=1


1


1


=x22x

33x

24






References














ndash 36 ndash

JHEP03(2016)165










(2003) 013 [hep-th0212208] [INSPIRE]




(2014) 302 [arXiv14036646] [INSPIRE]






















(2015) 006 [arXiv150208005] [INSPIRE]






ndash 37 ndash

JHEP03(2016)165



[INSPIRE]

















(2015) 29FT01 [arXiv150500963] [INSPIRE]








[INSPIRE]














ndash 38 ndash

JHEP03(2016)165















ndash 39 ndash

Introduction























JHEP03(2016)165


x22x23x

24

(part

partxmicro1+

part

partxmicro2+

part

partxmicro3+

part

partxmicro4

)[h

x22x23x

24

]= 0 (C4)


x1micropart

partx1νminus x1ν

part

partx1micro

)h= F1(Mmicroνe









x22x23x

24

(x1micro

part

partx1νminus x1ν

part


part

partx2νminus x2ν

part


part

partx3νminus x3ν

part

partx3micro

+x4micropart

partx4νminus x4ν

part

partx4micro

)[h

x22x23x

24

]= 0 (C6)


consider(x1 middot

part

partx1+ 3

)h = F1(e






x22x23x

24

(x1 middot

part

partx1+ x2 middot

part

partx2+ x3 middot

part

partx3+ x4 middot

part

partx4+ 12

)[h

x22x23x

24

]= 0 (C8)


h

x22x23x

24

=A




we have




ndash 34 ndash

JHEP03(2016)165





infinsumnm=0

im(minus1)n




where







minus (C14)

we find


and

infinsumn=0

(minus1)n




1


(C16)



=infinsum

n1n3n4=0

1



slot we get

α(P middot x2) =1


ndash 35 ndash

JHEP03(2016)165




=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)


=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)

(n1 + n3 + n4)



=infinsumn1

(minus1)n1




timesinfinsumn3

(minus1)n3




timesinfinsumn4

(minus1)n4





=1


1


1


=x22x

33x

24






References














ndash 36 ndash

JHEP03(2016)165










(2003) 013 [hep-th0212208] [INSPIRE]




(2014) 302 [arXiv14036646] [INSPIRE]






















(2015) 006 [arXiv150208005] [INSPIRE]






ndash 37 ndash

JHEP03(2016)165



[INSPIRE]

















(2015) 29FT01 [arXiv150500963] [INSPIRE]








[INSPIRE]














ndash 38 ndash

JHEP03(2016)165















ndash 39 ndash

Introduction























JHEP03(2016)165





infinsumnm=0

im(minus1)n




where







minus (C14)

we find


and

infinsumn=0

(minus1)n




1


(C16)



=infinsum

n1n3n4=0

1



slot we get

α(P middot x2) =1


ndash 35 ndash

JHEP03(2016)165




=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)


=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)

(n1 + n3 + n4)



=infinsumn1

(minus1)n1




timesinfinsumn3

(minus1)n3




timesinfinsumn4

(minus1)n4





=1


1


1


=x22x

33x

24






References














ndash 36 ndash

JHEP03(2016)165










(2003) 013 [hep-th0212208] [INSPIRE]




(2014) 302 [arXiv14036646] [INSPIRE]






















(2015) 006 [arXiv150208005] [INSPIRE]






ndash 37 ndash

JHEP03(2016)165



[INSPIRE]

















(2015) 29FT01 [arXiv150500963] [INSPIRE]








[INSPIRE]














ndash 38 ndash

JHEP03(2016)165















ndash 39 ndash

Introduction























JHEP03(2016)165




=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)


=infinsum

n1n3n4=0

1

n1n3n4(n1 + n3 + n4)

(n1 + n3 + n4)



=infinsumn1

(minus1)n1




timesinfinsumn3

(minus1)n3




timesinfinsumn4

(minus1)n4





=1


1


1


=x22x

33x

24






References














ndash 36 ndash

JHEP03(2016)165










(2003) 013 [hep-th0212208] [INSPIRE]




(2014) 302 [arXiv14036646] [INSPIRE]






















(2015) 006 [arXiv150208005] [INSPIRE]






ndash 37 ndash

JHEP03(2016)165



[INSPIRE]

















(2015) 29FT01 [arXiv150500963] [INSPIRE]








[INSPIRE]














ndash 38 ndash

JHEP03(2016)165















ndash 39 ndash

Introduction























JHEP03(2016)165










(2003) 013 [hep-th0212208] [INSPIRE]




(2014) 302 [arXiv14036646] [INSPIRE]






















(2015) 006 [arXiv150208005] [INSPIRE]






ndash 37 ndash

JHEP03(2016)165



[INSPIRE]

















(2015) 29FT01 [arXiv150500963] [INSPIRE]








[INSPIRE]














ndash 38 ndash

JHEP03(2016)165















ndash 39 ndash

Introduction























JHEP03(2016)165



[INSPIRE]

















(2015) 29FT01 [arXiv150500963] [INSPIRE]








[INSPIRE]














ndash 38 ndash

JHEP03(2016)165















ndash 39 ndash

Introduction























JHEP03(2016)165















ndash 39 ndash

Introduction























Published for SISSA by Springer2016)165.pdf · JHEP03(2016)165 Contents 1 Introduction1 2...

Documents

Transcript of Published for SISSA by Springer2016)165.pdf · JHEP03(2016)165 Contents 1 Introduction1 2...